What is it called when a large force acts on a body for a very short time interval the product of the force and time interval?

Isaac Newton's First Law of Motion states, "A body at rest will remain at rest, and a body in motion will remain in motion unless it is acted upon by an external force." What, then, happens to a body when an external force is applied to it? That situation is described by Newton's Second Law of Motion. 

According to NASA, this law states, "Force is equal to the change in momentum per change in time. For a constant mass, force equals mass times acceleration." This is written in mathematical form as F = ma

F is force, m is mass and a is acceleration. The math behind this is quite simple. If you double the force, you double the acceleration, but if you double the mass, you cut the acceleration in half. 

Newton published his laws of motion in 1687, in his seminal work "Philosophiæ Naturalis Principia Mathematica" (Mathematical Principles of Natural Philosophy) in which he formalized the description of how massive bodies move under the influence of external forces. 

Newton expanded upon the earlier work of Galileo Galilei, who developed the first accurate laws of motion for masses, according to Greg Bothun, a physics professor at the University of Oregon. Galileo's experiments showed that all bodies accelerate at the same rate regardless of size or mass. Newton also critiqued and expanded on the work of Rene Descartes, who also published a set of laws of nature in 1644, two years after Newton was born. Descartes' laws are very similar to Newton's first law of motion.

Newton's second law says that when a constant force acts on a massive body, it causes it to accelerate, i.e., to change its velocity, at a constant rate. In the simplest case, a force applied to an object at rest causes it to accelerate in the direction of the force. However, if the object is already in motion, or if this situation is viewed from a moving inertial reference frame, that body might appear to speed up, slow down, or change direction depending on the direction of the force and the directions that the object and reference frame are moving relative to each other.

The bold letters F and a in the equation indicate that force and acceleration are vector quantities, which means they have both magnitude and direction. The force can be a single force or it can be the combination of more than one force. In this case, we would write the equation as ∑F = ma

The large Σ (the Greek letter sigma) represents the vector sum of all the forces, or the net force, acting on a body. 

It is rather difficult to imagine applying a constant force to a body for an indefinite length of time. In most cases, forces can only be applied for a limited time, producing what is called impulse. For a massive body moving in an inertial reference frame without any other forces such as friction acting on it, a certain impulse will cause a certain change in its velocity. The body might speed up, slow down or change direction, after which, the body will continue moving at a new constant velocity (unless, of course, the impulse causes the body to stop).

There is one situation, however, in which we do encounter a constant force — the force due to gravitational acceleration, which causes massive bodies to exert a downward force on the Earth. In this case, the constant acceleration due to gravity is written as g, and Newton's Second Law becomes F = mg. Notice that in this case, F and g are not conventionally written as vectors, because they are always pointing in the same direction, down.

The product of mass times gravitational acceleration, mg, is known as weight, which is just another kind of force. Without gravity, a massive body has no weight, and without a massive body, gravity cannot produce a force. In order to overcome gravity and lift a massive body, you must produce an upward force ma that is greater than the downward gravitational force mg. 

Newton's second law in action

Rockets traveling through space encompass all three of Newton's laws of motion.

If the rocket needs to slow down, speed up, or change direction, a force is used to give it a push, typically coming from the engine. The amount of the force and the location where it is providing the push can change either or both the speed (the magnitude part of acceleration) and direction.

Now that we know how a massive body in an inertial reference frame behaves when it subjected to an outside force, such as how the engines creating the push maneuver the rocket, what happens to the body that is exerting that force? That situation is described by Newton’s Third Law of Motion. 

Additional reporting by Rachel Ross, Live Science contributor.

See also:

• Newton's Laws of Motion
• Inertia & Newton's First Law of Motion

Additional resources

As discussed previously, when two bodies collide, they exert large forces on one another (during the time of the collision) called impulsive forces. These forces are very large such that any other forces ( \(\mathrm {e}.\mathrm {g}\)., friction or gravity) present during the short time of the collision can be neglected. This approximation is known as the impulse approximation. For example, if a golf ball was hit by a golf club, the change in the momentum of the ball can be assumed to be only due to the impulsive force exerted on it by the club. The change in its momentum due to any other force present during the collision can be neglected. That is, the force in the expression I \(=\triangle \mathrm {p}=\overline{\mathrm {F}}\triangle t\) can be assumed to be the impulsive force only The neglected forces present during the collision time are external to the two-body system, whereas the impulsive forces are internal. The two-body system can therefore be considered to be isolated during the short time of the collision (which is in the order of a few milliseconds). Hence, the total linear momentum of the system is conserved during the collision, which enables us to apply the law of conservation of momentum immediately before and immediately after the collision. In general, for any type of collision, the total linear momentum is conserved during the time of the collision. That is, \(\mathrm {p}_{i}=\mathrm {p}_{f}\)., where \(\mathrm {p}_{i}\) and \(\mathrm {p}_{f}\) are the momenta immediately before and after the collision. In the next sections, we will define various types of two- body collisions, depending on whether or not the kinetic energy of the system is conserved.

Example 5.1

A 50 \(\mathrm {g}\) golf ball initially at rest is struck by a golf club. The golf club exerts a force on the ball that varies during a very short time interval from zero before impact, to a maximum value and back to zero when the ball is no longer in contact with the club. If the ball is given a speed of 25 \(\mathrm {m}/\mathrm {s}\), and if the club is in contact with the ball for \(7\times 10^{-}4\,\mathrm {s}\), find the average force exerted by the club on the ball.

Solution 5.1

The impulse of the force is

$$ I=\triangle p=mv_{f}-0=(0.05\,\mathrm {k}\mathrm {g})(25\,\mathrm {m}/\mathrm {s})=1.25\,\mathrm {k}\mathrm {g}\cdot \mathrm {m}/\mathrm {s} $$

the average force exerted on the ball by the club is then

$$ \overline{F}=\frac{I}{\triangle t}=\frac{(1.25\,\mathrm {k}\mathrm {g}\mathrm {m}/\mathrm {s})}{(7\times 10^{-4}\,\mathrm {s})}=1785.7\,\mathrm {N} $$

Example 5.2

A canon placed on a carriage fires a 250 kg ball to the horizontal with a speed of 50 \(\mathrm {m}/\mathrm {s}\). If the mass of the canon and the carriage is 4000 kg, find the recoil speed of the canon.

Solution 5.2

Because there are no external horizontal forces acting on the cannon-carriage-ball system, then the total momentum of the system is constant (conserved) in the \(\mathrm {x}\)-direction

$$ m_{1}v_{1f}+m_{2}v_{2f}=0 $$

therefore,

$$ v_{2f}=\frac{-m_{1}}{m_{2}}v_{1f}=-\frac{(250\,\mathrm {k}\mathrm {g})}{(4000\,\mathrm {k}\mathrm {g})}(50\,\mathrm {m}/\mathrm {s})=-3.1\,\mathrm {m}/\mathrm {s} $$

i.e., the cannon recoils in the negative \(\mathrm {x}\)-direction.

Example 5.3

A hockey puck of mass 0.16 kg traveling on a smooth ice surface collides with the court’s edge. If its initial and final velocities are \(\mathbf {v_{i}}=-2 \; \mathbf {i}\mathrm {m}/\mathrm {s}\) and \(\mathbf {v_{f}}=1 \; \mathbf {i}\mathrm {m}/\mathrm {s}\) and if the hockey puck is in contact with the wall for 2 ms, find the impulse delivered to the puck and the average force exerted on it by the wall.

Solution 5.3

$$\mathbf {I}=\triangle \mathbf {p}=\mathbf {p}_{f}-\mathbf {p}_{i}=m\mathbf {v}_{f}-m\mathbf {v}_{i}=(0.16\,\mathrm {k}\mathrm {g})((\mathrm {l}\,\mathrm {m}/\mathrm {s})-(-2\,\mathrm {m}/\mathrm {s}))\mathbf {i}=0.48\mathbf {i}\,\mathrm {k}\mathrm {g}\cdot \mathrm {m}/\mathrm {s}$$

$$ \overline{\mathbf {F}}=\frac{\mathbf {I}}{\triangle t}=\frac{(0.48 \; \mathbf {i}\,\mathrm {k}\mathrm {g}\cdot \mathrm {m}/\mathrm {s})}{(0.002\,\mathrm {s})}=240 \; \mathbf {i}\,\mathrm {N} $$

Example 5.4

A 0.5 kg hockey puck is initially moving in the negative \(\mathrm {y}\)-direction as shown in Fig. 5.3, with a speed of 7 \(\mathrm {m}/\mathrm {s}\). If a hockey player hits the puck giving it a velocity of magnitude 12 \(\mathrm {m}/\mathrm {s}\) in a direction of \(60^{\mathrm {o}}\) to the vertical, and if the collision lasts for 0.008 \(\mathrm {s}\), find the impulse due to the collision and the average force exerted on the puck.

A hockey player changing the momentum of the puck

Along the \(\mathrm {x}\)-direction, we have

and

$$p_{fx}=mv_{fx}=(0.5\,\mathrm {k}\mathrm {g})(12\mathrm {m}/\mathrm {s}) \cos 30^{o} =5.2\,\mathrm {k}\mathrm {g}\cdot \mathrm {m}/\mathrm {s}$$

along the \(\mathrm {y}\)-direction, we have

$$ p_{iy}=mv_{iy}=(0.5\,\mathrm {k}\mathrm {g})(-7\,\mathrm {m}/\mathrm {s})=-3.5\,\mathrm {k}\mathrm {g}\cdot \mathrm {m}/\mathrm {s} $$

and

$$p_{fy}=mv_{fy}=(0.5\,\mathrm {k}\mathrm {g})(12\,\mathrm {m}/\mathrm {s}) \sin 30^{\circ } =3\,\mathrm {k}\mathrm {g}\cdot \mathrm {m}/\mathrm {s}$$

Therefore, the impulse of the force in each direction is

$$ I_{x}=p_{fx}-p_{ix}=(5.2 \;\mathrm {k}\mathrm {g}\cdot \mathrm {m}/\mathrm {s})-0=5.2\,\mathrm {k}\mathrm {g}\cdot \mathrm {m}/\mathrm {s} $$

and

$$ I_{y}=p_{fy}-p_{iy}=(3\,\mathrm {k}\mathrm {g}\cdot \mathrm {m}/\mathrm {s})-(-3.5\,\mathrm {k}\mathrm {g}\cdot \mathrm {m}/\mathrm {s})=6.5\,\mathrm {k}\mathrm {g}\cdot \mathrm {m}/\mathrm {s} $$

$$ \mathbf {I}=(5.2 \; \mathbf {i}+6.5 \; \mathbf {j})\,\mathbf {k}\mathrm {g}\cdot \mathrm {m}/\mathrm {s} $$

$$ I=\sqrt{(5.2\,\mathrm {k}\mathrm {g}\mathrm {m}/\mathrm {s})^{2}+(6.5\,\mathrm {k}\mathrm {g}\mathrm {m}/\mathrm {s})^{2}}=8.3\,\mathrm {k}\mathrm {g}\cdot \mathrm {m}/\mathrm {s} $$

The direction of the impulse is

$$ \tan \theta =\frac{I_{y}}{I_{x}}=\frac{(6.5\,\mathrm {k}\mathrm {g}\cdot \mathrm {m}/\mathrm {s})}{(5.2\,\mathrm {k}\mathrm {g}\cdot \mathrm {m}/\mathrm {s})}=1.25 $$

$$ \theta =51.3^{\mathrm {o}} $$

where \(\theta \) is measured from the positive \(\mathrm {x}\)-axis. The average force acting on the puck is

$$ \overline{F}=\frac{I}{\triangle t}=\frac{(8.3\,\mathrm {k}\mathrm {g}\cdot \mathrm {m}/\mathrm {s})}{(0.008\,\mathrm {s})}=1037.5\,\mathrm {N} $$

Example 5.5

Two ice skaters of masses \(m_{1}=50\) kg and \(m_{1}=62\) kg standing face to face push each other on a frictionless horizontal surface. If skater (1) recoils with a speed of 5 \(\mathrm {m}/\mathrm {s}\), find the recoil speed of the other skater.

Solution 5.5

For the two-skater system, the sum of the vertical forces are zero (weight and normal forces) and the forces exerted by one skater on the other is internal to the system. That is, there are no external forces acting on the system and the total momentum is conserved. Because the motion takes place in a straight line, we have

$$ p_{1i}+p_{2i}=p_{1f}+p_{2f} $$

$$ 0=m_{1}v_{1f}+m_{2}v_{2f} $$

and hence,

$$ v_{2f}=\frac{-m_{1}}{m_{2}}v_{1f}=\frac{-(50\,\mathrm {k}\mathrm {g})}{(62\,\mathrm {k}\mathrm {g})}(5\,\mathrm {m}/\mathrm {s})=-4.03\,\mathrm {m}/\mathrm {s} $$

Example 5.6

A particle is moving in space under the influence of a force. If its momentum as a function of time is

$$ \mathbf {p}=((4t^{2}+t)\mathbf {i}-(3t-1)\mathbf {j}+(5t^{3}+2t)\mathbf {k})\ \mathbf {k}\mathrm {g}.\ \mathrm {m}/\mathrm {s} $$

(a) Find the force acting on the particle at any time; (b) Find the impulse of the force from \(t=0\) to \(t=1\,\mathrm {s}.\)

Solution 5.6

(a)

$$ \mathbf {F}=\frac{d\mathbf {p}}{dt}=((8t+1)\mathbf {i}-3\mathbf {j}+(15t^{2}+2)\mathbf {k})\,\mathrm {N} $$

(b)

$$ \mathbf {I}=\triangle \mathbf {p}=(5\mathbf {i}-2\mathbf {j}+7\mathbf {k})-\mathbf {j}=(5\mathbf {i}-3\mathbf {j}+7\mathbf {k})\, kg . \mathrm {m}/\mathrm {s} $$

5.4.1 Elastic Collisions

An elastic collision is one in which the total kinetic energy, as well as momentum, of the two-colliding-body system is conserved. These collisions exist when the impulsive force exerted by one body on the other is conservative. Such force converts the kinetic energy of the body into elastic potential energy when the two bodies are in contact. It then reconverts the elastic potential energy into kinetic energy when there is no more contact. After collision, each body may have a different velocity and therefore a different kinetic energy. However, the total energy as well as the total momentum of the system is constant during the time of the collision. An example of such collisions is those between billiard balls.

5.4.2 Inelastic Collisions

An inelastic collision is one in which the total kinetic energy of the two-colliding-body system is not conserved, although momentum is conserved. In such a collision, some of the kinetic energy of the system is lost due to deformation and appear as internal or thermal energy. In other words, the (internal) impulsive forces are not conservative. Therefore, the kinetic energy of the system before the collision is less than that after the collision. If the two colliding objects stick together, the collision is said to be perfectly inelastic. There are some types of collisions in which the total kinetic energy after the collision occurs is greater than that before it occurs. This type of collision is called an explosive collision.

Two particles of masses \(m_{1}\) and \(m_{2}\) experiencing an elastic head-on collision

When a collision takes place in one dimension, it is referred to as a head-on collision. Consider two particles of masses \(m_{1}\) and \(m_{2}\) experiencing an elastic head-on collision as in Fig. 5.4. Applying the law of conservation of energy and the law of conservation of linear momentum gives

$$ m_{1}\mathbf {v}_{1i}+m_{2}\mathbf {v}_{2i}=m_{1}\mathbf {v}_{1f}+m_{2}\mathbf {v}_{2f} $$

$$ \frac{1}{2}m_{1}v_{1i}^{2}+\frac{1}{2}m_{2}v_{2i}^{2}=\frac{1}{2}m_{1}v_{1f}^{2}+\frac{1}{2}m_{2}v_{2f}^{2} $$

Solving these equations for \(v_{1f}\) and \(v_{2f}\), we get

$$\begin{aligned} v_{1f}=\bigg (\displaystyle \frac{m_{1}-m_{2}}{m_{1}+m_{2}}\bigg )v_{1i}+\bigg (\frac{2m_{2}}{m_{1}+m_{2}}\bigg )v_{2i} \end{aligned}$$

(5.2)

$$\begin{aligned} v_{2f}=\bigg (\displaystyle \frac{2m_{1}}{m_{1}+m_{2}}\bigg )v_{1i}+\bigg (\frac{m_{2}-m_{1}}{m_{1}+m_{2}}\bigg )v_{2i} \end{aligned}$$

(5.3)

5.4.3.1 Special Cases

1. If \(m_{1}=m_{2}\), it follows from Eqs. 5.2 and 5.3 that \(v_{1f}=v_{2i}\) and \(v_{2f}= v_{1i}\). In other words, if the particles have equal masses they exchange velocities.

2. If \(m_{2}\) is stationary \((v_{2i}=0)\) , then from Eqs. 5.2 and 5.3, we have

$$\begin{aligned} v_{1f}=\bigg (\displaystyle \frac{m_{1}-m_{2}}{m_{1}+m_{2}}\bigg )v_{1i} \end{aligned}$$

(5.4)

$$\begin{aligned} v_{2f}=\bigg (\displaystyle \frac{2m_{1}}{m_{1}+m_{2}}\bigg )v_{1i} \end{aligned}$$

(5.5)

In that case \(m_{2}\) is called the target and \(m_{1}\) is called the projectile. Furthermore, if \(m_{1}\gg m_{2}\), then from Eqs. 5.4 and 5.5, we find that \(v_{1f}\approx v_{1i}\) and \(v_{2f}\approx 2v_{1i}\). While if \(m_{2}\gg m_{1}\), then from Eqs. 5.4 and 5.5, we see that \(v_{1f}\approx -v_{1i}\), and \(v_{2f}\approx v_{2i}=0.\)

A one dimensional (head-on) perfectly inelastic collision between two particles of mass \(m_{1}\) and \(m_{2}\)

Figure 5.5 shows a one-dimensional (head-on) perfectly inelastic collision between two particles of mass \(m_{1}\) and \(m_{2}\). Here, the kinetic energy of the system is not conserved, but the law of conservation of linear momentum still holds

$$ m_{1}\mathbf {v}_{1i}+m_{2}\mathbf {v}_{2i}=(m_{1}+m_{2})\mathbf {v}_{f} $$

$$ \mathbf {v}_{f}=\frac{m_{1}\mathbf {v}_{1i}+m_{2}\mathbf {v}_{2i}}{m_{1}+m_{2}} $$

5.4.5 Coefficient of Restitution

For any collision between two bodies in one dimension, the coefficient of restitution is defined as

$$ e=\frac{v_{2f}-v_{1f}}{v_{1i}-v_{2i}} $$

where \(v_{1i}\) and \(v_{2i}\) are velocities before the collision. \(v_{1f}\) and \(v_{2f}\) are velocities after the collision. \(|v_{1i}-v_{2i}|\) is called the relative speed of approach and \(|v_{2f}-v_{1f}|\) is the relative speed of recession.

• If \(e=1\) the collision is perfectly elastic.

• If \(e<1\) the collision is inelastic.

• If \(e=0\) the collision is perfectly inelastic (the two bodies stick together).

Example 5.7

Two marble balls of masses \(m_{1}=7\) kg and \(m_{2}=3\) kg are sliding toward each other on a straight frictionless track. If they experience a head-on elastic collision and if the initial velocities of \(m_{1}\) and \(m_{2}\) are 0.5 \(\mathrm {m}/\mathrm {s}\) to the right and 2 \(\mathrm {m}/\mathrm {s}\) to the left, respectively, find the final velocities of \(m_{1}\) and \(m_{2}.\)

Solution 5.7

For an elastic head-on collision, we have

$$ v_{1f}=\bigg (\displaystyle \frac{m_{1}-m_{2}}{m_{1}+m_{2}}\bigg )v_{1i}+\bigg (\frac{2m_{2}}{m_{1}+m_{2}}\bigg )v_{2i}=(0.4)(0.5\,\mathrm {m}/\mathrm {s})+(0.6)(-2\,\mathrm {m}/\mathrm {s})=-1\,\mathrm {m}/\mathrm {s} $$

$$ v_{2f}=\bigg (\displaystyle \frac{2m_{1}}{m_{1}+m_{2}}\bigg )v_{1i}+\bigg (\frac{m_{2}-m_{1}}{m_{1}+m_{2}}\bigg )v_{2i}=(1.4)(0.5\,\mathrm {m}/\mathrm {s})+(-0.4)(-2\,\mathrm {m}/\mathrm {s})=1.5\,\mathrm {m}/\mathrm {s} $$

Example 5.8

The ballistic pendulum consists of a large wooden block suspended by a light wire (see Fig. 5.6). The system is used to measure the speed of a bullet where the bullet is fired horizontally into the block. The collision is perfectly inelastic and the system (bullet\(+\)block) swings up a height h. If \(M=3\) kg, \(m=5\,\mathrm {g}\) and \(h=5\) cm, find (a) the initial speed of the bullet; (b) the mechanical energy lost due to the collision.

The ballistic pendulum consists of a large wooden block suspended by a light wire

(a) Using the impulse approximation, the law of conservation of momentum gives the velocities just before and after the collision when the string is still nearly vertical. For a perfectly inelastic collision, the total momentum is conserved but the total kinetic energy is not conserved during the collision. Thus, we have

$$ v_{1i}=\frac{(m+M)}{m}v_{f} $$

After the collision, the energy of the (bullet \(+\mathrm {b}\mathrm {l}\mathrm {o}\mathrm {c}\mathrm {k}+\)earth) system is conserved since the gravitational force is the only force acting in the system.

$$ \frac{1}{2}(m+M)v_{f}^{2}=(m+M)gh $$

That gives

$$ v_{1i}=\frac{(m+M)}{m}\sqrt{2gh}=\frac{(3.005 \; \mathrm {k}\mathrm {g})}{(0.005\,\mathrm {k}\mathrm {g})}\sqrt{2(9.8 \,\mathrm {m}/\mathrm {s}^{2})(0.05\,\mathrm {m})}=595 \; \mathrm {m}/\mathrm {s} $$

(b) The kinetic energy of the bullet before collision is

$$ \displaystyle \frac{1}{2}mv_{1i}^{2}=\frac{1}{2} 0.005 \; \mathrm {k}\mathrm {g})(595\,\mathrm {m}/\mathrm {s})^{2}=885 \; \mathrm {J} $$

After collision, the kinetic energy of the (bullet\(+\)block) is

$$ \frac{1}{2}(m+M)v_{f}^{2}=(m+M)\ (gh)=(3.005\,\mathrm {k}\mathrm {g})(9.8\,\mathrm {m}/\mathrm {s}^{2})(0.05\,\mathrm {m})=1.5 \; \mathrm {J} $$

therefore,

$$ \triangle E=(885 \; \mathrm {J})-(1.5 \; \mathrm {J})=883.5 \; \mathrm {J} $$

That is, nearly, all the mechanical energy is dissipated and converted into internal (thermal) energy of the (block\(+\)bullet) system.

Example 5.9

Two masses \(m_{1}=0.8\) kg and \(m_{2}=0.5\) kg are heading toward each other with speeds of 0.25 \(\mathrm {m}/\mathrm {s}\) and \(-0.5\,\mathrm {m}/\mathrm {s}\), respectively. If they have a perfectly inelastic collision, find the final velocity of the system just after the collision.

Solution 5.9

$$v_{f}=\displaystyle \frac{m_{1}v_{1i}+m_{2}v_{2i}}{(m_{1}+m_{2})}=\frac{(0.8\,\mathrm {k}\mathrm {g})(0.25\,\mathrm {m}/\mathrm {s})-(0.5\,\mathrm {k}\mathrm {g})(0.5\,\mathrm {m}/\mathrm {s})}{(1.3\,\mathrm {k}\mathrm {g})}=-0.04\,\mathrm {m}/\mathrm {s}$$

Example 5.10

Two blocks \(m_{1}=2\) kg and \(m_{2}=1\) kg collide head-on with each other on a frictionless surface (see Fig. 5.7. If \(v_{1i}=-10\,\mathrm {m}/\mathrm {s}\) and \(v_{2i}=15\,\mathrm {m}/\mathrm {s}\) and the coefficient of restitution is \(e=1/4\), determine the final velocities of the masses just after the collision.

Two blocks colliding head-on on a frictionless surface

$$ e=\frac{v_{2f}-v_{1f}}{v_{1i}-v_{2i}} $$

$$ \frac{1}{4}=\frac{v_{2f}-v_{1f}}{(-25\mathrm {m}/\mathrm {s})} $$

$$\begin{aligned} v_{2f}-v_{1f}=-6.25\,\mathrm {m}/\mathrm {s} \end{aligned}$$

(5.6)

From the conservation of momentum, we have

$$ m_{1}v_{1i}+m_{2}v_{2i}=m_{1}v_{1f}+m_{2}v_{2f} $$

$$ (2\,\mathrm {k}\mathrm {g})(-10\,\mathrm {m}/\mathrm {s})+(1\,\mathrm {k}\mathrm {g})(15\,\mathrm {m}/\mathrm {s})=(2\,\mathrm {k}\mathrm {g})v_{1f}+(1\,\mathrm {k}\mathrm {g})v_{2f} $$

That gives

$$\begin{aligned} v_{2f}+(2\,\mathrm {k}\mathrm {g})v_{1f}=-5\,\mathrm {m}/\mathrm {s} \end{aligned}$$

(5.7)

Solving Eqs. 5.6 and 5.7 gives \(v_{1f}=0.42 \,\mathrm {m}/\mathrm {s}\) and \(v_{2f}=-5.83 \; \mathrm {m}/\mathrm {s}.\)

Example 5.11

A \(m_{1}=5\,\mathrm {g}\) bullet is fired horizontally at the center of a wooden block with a mass of \(m_{2}=2\,\mathrm {k}\mathrm {g}\). The bullet embeds itself in the block and the two slides a distance of 0.\(5\,\mathrm {m}\) on a rough surface \((\mu _{k}=0.2)\) before coming to rest. Find the initial speed of the bullet.

Solution 5.11

Applying the law of conservation of momentum immediately before and after the collision gives

$$ m_{1}v_{1i}+0=(m_{1}+m_{2})v_{f} $$

$$ v_{1i}=\frac{(2.005\,\mathrm {k}\mathrm {g})}{(0.005 \; \mathrm {k}\mathrm {g})}v_{f}=(401)v_{f} $$

by taking the (block\(+\)bullet) as the system after the collision until it comes to rest, we have

$$ K_{f}+U_{f}=K_{i}+U_{i}+\triangle K_{ext} $$

that gives

$$ 0=\frac{1}{2}(m_{1}+m_{2})v_{f}^{2}-\mu _{k}(m_{1}+m_{2})gd $$

$$ v_{f}=\sqrt{2\mu _{k}gd}=\sqrt{2(0.2)(9.8\,\mathrm {m}/\mathrm {s}^{2})(0.5\,\mathrm {m})}=1.4\,\mathrm {m}/\mathrm {s} $$

Hence,

$$ v_{1i}=(401)(1.4\,\mathrm {m}/\mathrm {s})=561.4\,\mathrm {m}/\mathrm {s} $$

5.4.6 Collision in Two Dimension

When a collision takes place in space, the total linear momentum is conserved along each of the \(x-, y\)-, and z-directions. That is, \(p_{ix}=p_{fx}, p_{iy}=p_{fy}\), and \(p_{iz}=p_{fz}.\) Here, we will analyze a two-dimensional elastic collision between two particles where one particle is moving and the other is at rest as shown in Fig. 5.8. This type of collision is known as a glancing collision. Since the collision is elastic, it follows that the total linear momentum as well as the kinetic energy of the system are conserved. Applying these laws immediately before and immediately after the collision, we have \(p_{ix}=p_{fx}\) and \(p_{iy}=p_{fy}\) or

$$ m_{1}v_{1ix}+m_{2}v_{2ix}=m_{1}v_{1fx}+m_{2}v_{2fx} $$

and

$$ m_{1}v_{1iy}+m_{2}v_{2iy}=m_{1}v_{1fy}+m_{2}v_{2fy} $$

From Fig. 5.8, we have

$$ m_{1}v_{1i}=m_{1}v_{1f}\cos \alpha _{1}+m_{2}v_{2f}\cos \alpha _{2} $$

and

$$ 0=m_{1}v_{1f}\sin \alpha _{1}+m_{2}v_{2f}\sin \alpha _{2} $$

Furthermore,

$$ \frac{1}{2}m_{1}v_{1i}^{2}=\frac{1}{2}m_{1}v_{1f}^{2}+\frac{1}{2}m_{2}v_{2f}^{2} $$

Therefore, we have three equations and seven unknown quantities. By knowing any four of these quantities, the three equations for the three variables can be solved.

A two dimensional elastic collision between two particles where one particle is moving and the other is at rest

A ball of mass of 2 kg is sliding along a horizontal frictionless surface at a speed of 3 \(\mathrm {m}/\mathrm {s}\). It then collides with a second ball of mass of 5 kg that is initially at rest. After the collision, the second ball is deflected with a speed of 1 \(\mathrm {m}/\mathrm {s}\) at an angle of \(30^{\mathrm {o}}\) below the horizontal as shown in Fig. 5.9. (a) Find the final velocity of the first ball; (b) show that the collision is inelastic; (c) suppose that the two balls have equal masses and the collision is perfectly elastic, show that \(\theta _{1}+\theta _{2}=90^{\mathrm {o}}.\)

A ball sliding along a horizontal frictionless surface collides with a second ball that is initially at rest

Applying the law of conservation of momentum immediately before and after the collision in each direction gives \(p_{ix}=p_{fx}\) and \(p_{iy}=p_{fy}\). Thus,

$$ m_{1}v_{1ix}+m_{2}v_{2ix}=m_{1}v_{1fx}+m_{2}v_{2fx} $$

$$v_{1fx}=\displaystyle \frac{m_{1}v_{1ix}+m_{2}v_{2ix}-m_{2}v_{2fx}}{m_{1}}=\frac{(2\,\mathrm {k}\mathrm {g})(3\,\mathrm {m}/\mathrm {s})+0-((5\,\mathrm {k}\mathrm {g})(1\,\mathrm {m}/\mathrm {s})\cos (-30))}{(2\,\mathrm {k}\mathrm {g})}$$

$$ v_{1fx}=0.84\,\mathrm {m}/\mathrm {s} $$

Along the \(\mathrm {y}\)-direction, we have

$$ m_{1}v_{1iy}+m_{2}v_{2iy}=m_{1}v_{1fy}+m_{2}v_{2fy} $$

$$ v_{1fy}=\frac{m_{1}v_{1iy}+m_{2}v_{2iy}-m_{2}v_{2fy}}{m_{1}}=\frac{0-((5\,\mathrm {k}\mathrm {g})(1\,\mathrm {m}/\mathrm {s})\sin (-30^{\mathrm {o}}))}{(2\,\mathrm {k}\mathrm {g})} $$

$$ v_{1fy}=1.25\,\mathrm {m}/\mathrm {s} $$

Thus, the final velocity of the first ball is

$$ v_{1f}=\sqrt{v_{1fx}^{2}+v_{1fy}^{2}}=\sqrt{(0.84\,\mathrm {m}/\mathrm {s})^{2}+(1.25\,\mathrm {m}/\mathrm {s})^{2}}=1.5\,\mathrm {m}/\mathrm {s} $$

The direction of the velocity is

$$ \tan \theta _{1}=\frac{v_{1fy}}{v_{1fx}}=\frac{(1.25\,\mathrm {m}/\mathrm {s})}{(0.84\,\mathrm {m}/\mathrm {s})}=1.5 $$

$$ \theta _{1}=56^{\mathrm {o}} $$

(b) The total kinetic energy before the collision is

$$ K_{i}=\frac{1}{2}m_{1}v_{1i}^{2}=\frac{1}{2}(2\,\mathrm {k}\mathrm {g})(3\,\mathrm {m}/\mathrm {s})^{2}=9\,\mathrm {J} $$

The total kinetic energy after the collision is

$$K_{f}=\displaystyle \frac{1}{2}m_{1}v_{1f}^{2}+\frac{1}{2}m_{2}v_{2f}^{2}=\frac{1}{2}(2\,\mathrm {k}\mathrm {g})(1.5\,\mathrm {m}/\mathrm {s})^{2}+\frac{1}{2}(5\,\mathrm {k}\mathrm {g})(1\,\mathrm {m}/\mathrm {s})^{2}=4.75 \; \mathrm {J}$$

That is, some of the energy of the system is lost and thus the collision is inelastic.

(c) In a perfectly elastic collision, both the total momentum and the total mechanical energy of the system are conserved. That is

$$ m_{1}v_{1ix}+m_{2}v_{2ix}=m_{1}v_{1fx}+m_{2}v_{2fx} $$

$$\begin{aligned} v_{1i}=v_{1f}\cos \theta _{1}+v_{2f}\cos \theta _{2} \end{aligned}$$

(5.8)

$$ 0=v_{1f}\sin \theta _{1}-v_{2f}\sin \theta _{2} $$

$$\begin{aligned} v_{1f}\sin \theta _{1}=v_{2f}\sin \theta _{2} \end{aligned}$$

(5.9)

From the conservation of kinetic energy, we have

$$ \frac{1}{2}m_{1}v_{1i}^{2}=\frac{1}{2}m_{1}v_{1f}^{2}+\frac{1}{2}m_{2}v_{2f}^{2} $$

or

$$\begin{aligned} v_{1i}^{2}=v_{1f}^{2}+v_{2f}^{2} \end{aligned}$$

(5.10)

Substituting Eq. 5.8 into Eq. 5.9 gives

$$v_{1i}=v_{2f}\displaystyle \frac{\sin \theta _{2}}{\sin \theta _{1}} \cos \theta _{1}+v_{2f} \cos \theta _{2}$$

or

$$\begin{aligned} v_{1i}=\displaystyle \frac{v_{2f}\sin (\theta _{1}+\theta _{2})}{\sin \theta _{1}} \end{aligned}$$

(5.11)

Substituting Eq. 5.11 into Eq. 5.10 gives

$$ \frac{v_{2f}^{2}\sin ^{2}(\theta _{1}+\theta _{2})}{\sin ^{2}\theta _{1}}=\frac{v_{2f}^{2}\sin ^{2}\theta _{2}}{\sin ^{2}\theta _{1}}+v_{2f}^{2} $$

Therefore,

$$ \sin ^{2}(\theta _{1}+\theta _{2})=\sin ^{2}\theta _{1}+\sin ^{2}\theta _{2} $$

This is satisfied only if \(\theta _{1}+\theta _{2}=90^{\mathrm {o}}\).

A 1200 kg car traveling east at a speed of 18 \(\mathrm {m}/\mathrm {s}\) collides with another car of mass of 2500 kg that is traveling north at a speed of 23 \(\mathrm {m}/\mathrm {s}\)

A 1200 kg car traveling east at a speed of 18 \(\mathrm {m}/\mathrm {s}\) collides with another car of mass of 2500 kg that is traveling north at a speed of 23 \(\mathrm {m}/\mathrm {s}\) as shown in Fig. 5.10. If the collision is perfectly inelastic, how much mechanical energy is lost due to the collision?

Solution 5.13

$$ m_{1}v_{1ix}=(m_{1}+m_{2})v_{fx} $$

$$ v_{fx}=\frac{m_{1}v_{1ix}}{(m_{1}+m_{2})}=\frac{(1200\,\mathrm {k}\mathrm {g})(18\,\mathrm {m}/\mathrm {s})}{(3700\,\mathrm {k}\mathrm {g})}=5.8\,\mathrm {m}/\mathrm {s} $$

$$ m_{2}v_{2iy}=(m_{1}+m_{2})v_{fy} $$

$$ v_{fy}=\frac{m_{2}v_{2iy}}{(m_{1}+m_{2})}=\frac{(2500\,\mathrm {k}\mathrm {g})(23\,\mathrm {m}/\mathrm {s})}{(3700\,\mathrm {k}\mathrm {g})}=15.5\,\mathrm {m}/\mathrm {s} $$

$$ v_{f}=\sqrt{v_{fx}^{2}+v_{fy}^{2}}=\sqrt{(5.8\,\mathrm {m}/\mathrm {s})^{2}+(15.5\,\mathrm {m}/\mathrm {s})^{2}}=16.5\,\mathrm {m}/\mathrm {s} $$

The direction of \(v_{f}\) is

$$ \theta =\tan ^{-1}\frac{v_{fy}}{v_{fx}}=\tan ^{-1}\frac{(15.5\,\mathrm {m}/\mathrm {s})}{(5.8\,\mathrm {m}/\mathrm {s})}=69.5^{\mathrm {o}} $$

from the positive \(\mathrm {x}\)-axis. The change in the kinetic energy of the system is

$$ \triangle K=K_{f}-K_{i}=\frac{1}{2}(m_{1}+m_{2})v_{f}^{2}-\bigg (\frac{1}{2}m_{1}v_{1i}^{2}+\frac{1}{2}m_{2}v_{2i}^{2}\bigg ) $$

$$=\displaystyle \frac{1}{2} (3700 \; \displaystyle \mathrm {k}\mathrm {g})(16.5\,\mathrm {m}/\mathrm {s})^{2}-\bigg (\frac{1}{2} (1200 \; \displaystyle \mathrm {k}\mathrm {g})(18\,\mathrm {m}/\mathrm {s})^{2}+\frac{1}{2} (2500\,\mathrm {k}\mathrm {g})(23\,\mathrm {m}/\mathrm {s})^{2}\bigg )$$

$$ \triangle K=-3.5\times 10^{5} \; \mathrm {J} $$

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An isolated system consisting of two particles where the only forces that act in the system are internal forces