# Can I use $v_{final}=0$ when I calculate impulsive force?

If a 1kg ball strikes a wall with 10m/s and it rebounds with 10m/s, the time of impact being 0.2s, what is the impulsive force? I know that the initial velocity is 10m/s but can I state that the final velocity is zero because what I want to calculate is the velocity when the ball touches the wall. If we assume that there is no energy loss to the surrounding,the answer we get is 100N, but if there is energy loss to the surrounding, will I get the same answer as well?

• If there is energy loss you will have an inelastic collision. Sep 27 '18 at 15:29
• Carefully read the chapter in your physics book regarding momentum and impulse. Then, realize that the velocity going into the wall is positive and the velocity upon rebound is negative. Sep 27 '18 at 15:41
• If there was energy loss, rebounce velocity would not be $10\frac{m}{s}$. You can then recalculate the impact force in the same way, taking into consideration the new final velocity. Sep 27 '18 at 15:49

When you calculate the force on the ball using the change in momentum of the ball and the time taken, you have calculated the average force.

If you calculate the average force when the velocity of the ball becomes zero(ie after 0.1s), and when the ball has rebounded and left contact with the wall again (after 0.2s), then you will get the same answer of 100N. The change in velocity is only 5m/s and the time is 0.1s, compared with 10m/s and 0.2s. But the ratio is the same, so the average force is the same.

However, the half-way stage in a perfectly elastic collision is not the same as a perfectly inelastic collision. In both cases the ball has stopped moving, but in the inelastic collision it has stopped moving permanently, whereas in the elastic case it has stopped moving temporarily. In the elastic case there is elastic energy stored in the deformed ball which is waiting to be released; in the inelastic case there is no stored energy - it has already been dissipated.

It is impossible to deduce from the average force in a perfectly elastic collision what the average force would be in a perfectly inelastic collision with the same mass of ball and initial velocity. The reason is that it is a completely different situation. The material which the ball is made of must be different, otherwise it would behave the same way (ie elastically). So the duration of the collision is not guaranteed to be half that of the elastic collision. It could be more than half, it could be less. There is no way to predict what it would be.

Neither can you deduce that all elastic collisions of a ball of mass 1kg striking the wall with a speed of 10m/s will result in an average force of 100N. Balls and walls which are made of different elastic materials (eg rubber, steel) will take different times to collide, so the average force will be different.

Conclusion : Force and elasticity are not related.

I’m not sure what you meant by the title of the question (Can I use v = 0 when calculate impulsive force?), but if there is energy lost to the surroundings you will not get the same impulse force.

If there is energy lost (not lost, but dissipated as heat in the wall and/or ball), the rebound kinetic energy will be less, and therefore the rebound velocity will be less. If you do the impulse of force calculation with a smaller rebound velocity (make sure you keep the signs of the velocity correct), you will obtain a lower force.

Hope this helps