0.2 Force, momentum and impulse  (Page 27/35)

Exercise

1. Which one of the following is NOT a unit of impulse?
   1. A $N·s$
   2. B $kg·\mathrm{m}·{\mathrm{s}}^{-1}$
   3. C $J·\mathrm{m}·{\mathrm{s}}^{-1}$
   4. D $J·m^{-1}·s$
2. A toy car of mass 1 kg moves eastwards with a speed of 2 m $·$ s ${}^{-1}$ . It collides head-on with a toy train. The train has a mass of 2 kg and is moving at a speed of 1,5 m $·$ s ${}^{-1}$ westwards. The car rebounds (bounces back) at 3,4 m $·$ s ${}^{-1}$ and the train rebounds at 1,2 m $·$ s ${}^{-1}$ .
   1. Calculate the change in momentum for each toy.
   2. Determine the impulse for each toy.
   3. Determine the duration of the collision if the magnitude of the force exerted by each toy is 8 N.
3. A bullet of mass 20 g strikes a target at 300 m $·$ s ${}^{-1}$ and exits at 200 m $·$ s ${}^{-1}$ . The tip of the bullet takes 0,0001s to pass through the target. Determine:
   1. the change of momentum of the bullet.
   2. the impulse of the bullet.
   3. the magnitude of the force experienced by the bullet.
4. A bullet of mass 20 g strikes a target at 300 m $·$ s ${}^{-1}$ . Determine under which circumstances the bullet experiences the greatest change in momentum, and hence impulse:
   1. When the bullet exits the target at 200 m $·$ s ${}^{-1}$ .
   2. When the bullet stops in the target.
   3. When the bullet rebounds at 200 m $·$ s ${}^{-1}$ .
5. A ball with a mass of 200 g strikes a wall at right angles at a velocity of 12 m $·$ s ${}^{-1}$ and rebounds at a velocity of 9 m $·$ s ${}^{-1}$ .
   1. Calculate the change in the momentum of the ball.
   2. What is the impulse of the wall on the ball?
   3. Calculate the magnitude of the force exerted by the wall on the ball if the collision takes 0,02s.
6. If the ball in the previous problem is replaced with a piece of clay of 200 g which is thrown against the wall with the same velocity, but then sticks to the wall, calculate:
   1. The impulse of the clay on the wall.
   2. The force exerted by the clay on the wall if it is in contact with the wall for 0,5 s before it comes to rest.

Conservation of momentum

In the absence of an external force acting on a system, momentum is conserved.

Conservation of Linear Momentum

The total linear momentum of an isolated system is constant. An isolated system has no forces acting on it from the outside.

This means that in an isolated system the total momentum before a collision or explosion is equal to the total momentum after the collision or explosion.

Consider a simple collision of two billiard balls. The balls are rolling on a frictionless surface and the system is isolated. So, we can apply conservation of momentum. The first ball has a mass $m_1$ and an initial velocity $v_{i1}$ . The second ball has a mass $m_2$ and moves towards the first ball with an initial velocity $v_{i2}$ . This situation is shown in [link] .

The total momentum of the system before the collision, $p_i$ is:

After the two balls collide and move away they each have a different momentum. If the first ball has a final velocity of $v_{f1}$ and the second ball has a final velocity of $v_{f2}$ then we have the situation shown in [link] .

The total momentum of the system after the collision, $p_f$ is:

This system of two balls is isolated since there are no external forces acting on the balls. Therefore, by the principle of conservation of linear momentum, the total momentum before the collision is equal to the total momentum after the collision. This gives the equation for the conservation of momentum in a collision of two objects,