Question

A 2kg bowling ball exerts a 10n force on a wall. after the collision the ball is going -6m/s. if the collision last 0.03s, what was the initial velocity of the ball?

Answer 1

The initial velocity of the 2kg bowling ball was 6.15m/s directed towards the wall before collision.

Step-by-step explanation:

To determine the initial velocity of a 2kg bowling ball that exerts a 10N force on a wall and moves at -6m/s after collision, we can use the impulse-momentum theorem. The impulse exerted on an object is equal to the change in momentum of the object, with impulse being the product of the force and the time of contact (impulse=force×time) and momentum being the product of mass and velocity (momentum = mass× velocity).

The final velocity (Vf) is given as -6m/s. Since the ball is coming back after the collision, the velocity is in the opposite direction of the initial velocity. The time of collision is given as 0.03s.

First, we calculate the impulse: Impulse = 10N × 0.03s = 0.3 N×s

Then, we calculate the change in momentum:
Change in momentum = mass × change in velocity = mass × (Vf - Vi)

Since Impulse = Change in momentum, 0.3 N×s = 2kg × (-6m/s - Vi)

Solving for Vi we get: Vi = -6m/s - (0.3 N×s / 2kg) = -6m/s - 0.15m/s = -6.15m/s

The negative sign indicates that the initial velocity was in the opposite direction to the final velocity after the collision, which means that the ball was traveling towards the wall at 6.15m/s before hitting the wall.