Question

Darna rolls the 7.05-kg ball down the lane and it hits the 1.52-kg pin head on. The ball was moving at 8.24 m/s before the collision. The pin went flying forward at 13.2 m/s. Determine the post-collision speed of the ball.

Answer 1

Answer:

The velocity of the ball after the collision is 5.39 m/s

Explanation:

Hi there!

To solve this problem, we will use the conservation of momentum: the momentum of the system ball-pin remains the same before and after the collision. The momentum of the system is calculated as follows:

momentum before the collision (initial momentum) = mb · vb1 + mp · vp1

momentum after the collision (final momentum) = mb · vb2 + mp · vp2

Where:

mb = mass of the ball = 7.05 kg

vb1 = velocity of the ball before the collision = 8.24 m/s

mp = mass of the pin = 1.52 kg

vp1 = velocity of the pin before the collision = 0 m/s.

vb2 = velocity of the ball after the collsion = unknown.

vp2 = velocity of the pin after the collision = 13.2 m/s

Since momentum is conserved, then:

initial momentum = final momentum

mb · vb1 + mp · vp1 = mb · vb2 + mp · vp2

Solving for vb2:

mb · vb1 + mp · vp1 -  mp · vp2 = mb · vb2

(mb · vb1 + mp · vp1 -  mp · vp2) / mb = vb2

Since the pin is initially at rest, vp1 = 0:

(mb · vb1  -  mp · vp2) / mb = vb2

(7.05 kg · 8.24 m/s - 1.52 kg · 13.2 m/s) / 7.05 kg = vb2

vb2 = 5.39 m/s

The velocity of the ball after the collision is 5.39 m/s