A 1.6-kg ball is attached to the end of a 0.40-m string to form a pendulum. This pendulum is released from rest with the string horizontal. At the lowest point of its swing, when it is moving horizontally, the ball collides with a 0.80-kg block initially at rest on a horizontal frictionless surface. The speed of the block just after the collision is 3.0 m/s. What is the speed of the ball just after the collision?
A) 1.7 m/s
B) 1.1 m/s
C) 1.5 m/s
D) 1.3 m/s
E) 2.1 m/s

Question

Quizplus · Accepted Answer

Conservation of momentum can be applied to both the ball and the block before and after collision. Let's assume the velocity of the ball just after the collision to be v.

Before collision:
Momentum of ball = 0 (since it is at rest)
Momentum of block = 0 (since it is at rest)

Total momentum = 0

After collision:
Momentum of ball = mv
Momentum of block = 0.8 x 3 = 2.4

Total momentum = mv + 2.4

Since momentum is conserved, we can equate the total momentum before and after the collision:

0 = mv + 2.4

Solving for v, we get:

v = -2.4/m = -2.4/1.6 = -1.5 m/s

The negative sign indicates that the ball is moving in the opposite direction to its original direction of motion. To get the magnitude of velocity, we take the absolute value:

|v| = 1.5 m/s

Therefore, the speed of the ball just after the collision is 1.3 m/s (Option D).

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