Question

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 impulses are the same.

B. The second collision.

C. The first collision.

D. Cannot be determined without knowing the mass of the cart.

E. Cannot be determined without knowing the rebound speed of the first collision.

Answer 1

The impulse on the cart is greater in the first collision, where the change in velocity is greater.

Step-by-step explanation:

In this given situation, we have two different collisions: the first collision where the cart rebounds in the opposite direction and the second collision where the cart is at rest after the collision. The impulse on an object is equal to the change in momentum. Since momentum is equal to mass times velocity, the impulse is directly proportional to the change in velocity.

In the first collision, the cart changes its velocity from 0.3 m/s to -0.3 m/s, resulting in a change in velocity of 0.6 m/s. In the second collision, the cart changes its velocity from 0.3 m/s to 0 m/s, resulting in a change in velocity of 0.3 m/s.

Therefore, the impulse on the cart is greater in the first collision, where the change in velocity is greater.

Answer 2

C. In the first collision has twice the momentum as when it stays still ( second colllions)

Step-by-step explanation:

To see which statement is correct, it is best to solve the problem, the momentum is equal to the variation of the moment

I = Δp = m vf - m v₀

I = m (vf -v₀)

Case 1. In car bounces, the initial speed is 0.3 m / s, say that this direction is positive, when the magnitude of the speed bounces it remains constant, but its direction is reversed (vf = -0.3 m / s)

I₁ = m (-0.3 - 0.3)

I₁ = -0.6 m

Case 2. The expensive one that still after the crash so its speed is zero (vf = 0)

I₂ = m (0 - 0.3)

I₂ = -0.3 m

Let's calculate the relationship between the two impulses

I₁ / I₂ = -0.6m / -0.3m

I₂ / I₂ = 2

When it bounces it has twice the momentum as when it stays still

Now let's analyze the answers:

A. False The momentum changes

B. False. The momentum is less in the second collision

C. True. The momentum is double in this collision

D. False. Can be calculated, because the mass is the same throughout the exercise and is eliminated in the equations

E. False. When they say bounces it implies the same speed with the opposite direction