A cart traveling at 0.3 m/s collides with stationary object. After the collision, the cart rebounds in the opposite direction. The same cart again traveling at 0.3 m/s collides with a different stationary object. This time the cart is at rest after the collision. In which collision is the impulse on the cart greater?A. The first collision. B. Cannot be determined without knowing the rebound speed of the first collision. C. The second collision. D. Cannot be determined without knowing the mass of the cart. E. The impulses are the same.

Explanation:

First case

A cart speed is 0.3m/s. i.e the initial velocity is u=0.3m/s

It collide with a stationary body, then after collision the ball rebounds and move in opposite direction. This shows that the ball have a velocity after impulse let say v

Then, impulse is given as the change in linear momentum of a body

Impulse =m∆v

I=m(v-u)

Note, momentum is a vector quantity.

I=m(v--u)

I=m(v+u)

I=m(v+0.3)

I¹=0.3m+mv. Equation 1

Second case

A cart speed is 0.3m/s. i.e the initial velocity is u=0.3m/s

It collide with a stationary body, then after collision the ball is at rest, this show that the final velocity is v=0

Then, impulse is given as the change in linear momentum of a body

Impulse =m∆v

I=m(v-u)

Note, momentum is a vector quantity.

I=m(v--u)

I=m(v+u)

In this case v=0 u=0.3m/s

I=m(0+0.3)

I²=0.3m. Equation 2

If we compare impulse 1 (I¹) to impulse 2 (I²)

Subtract equation 2 from 1

We have, I¹ - I² =0.3m+mv -0.3m

I¹ - I² =mv

I¹ =mv+I²

We notice that the first impulse (I¹) is greater than second impulse (I²) by mv.

The correct answer is A

f=ma a/f/m a= 480/60 a = 8m/s/s that is quick wow