A 140 g block on a frictionless table is firmly attached to one end of a spring

Question

A 140 g block on a frictionless table is firmly attached to one end of a spring with k = 30 N/m. The other end of the spring is anchored to the wall. A 25 g ball is thrown horizontally toward the block with a speed of 5.8 m/s.

Part A) If the collision is perfectly elastic, what is the ball's speed immediately after the collision?

Part B) What is the maximum compression of the spring?

Part C) Repeat part A for the case of a perfectly inelastic collision.

Part D) Repeat part B for the case of a perfectly inelastic collision.

Explanation / Answer

The mass of block, m1 = 140 g = 0.14 kg

The initial velocity of m1, u1= 0

The mass of ball, m2 = 0.025 kg

initial veloctiy of m2, u2 = 5.8 m/s

Spring constant, k = 30 N/m

Part A:

If the collision is perfectly elastic,

the ball speed after collision, v2 = [ 2m1u1 + (m2-m1)u2 ] / (m1 +m2)

v2 = -4.04 m/s

Part B:

After collision, the speed of the block is

v1 = [ 2m2u2 + (m1-m2)u1 ] / (m1 +m2) = 1.76 m/s

We have, (1/2)m1v1^2 = (1/2)kx^2

x = [m1v1^2 / k]^(1/2) = 0.12 m

So the compression in the spring, x = 0.051 m

Part C:

If the collision is perfectly in elastic collision,

The velocity of block after collision, v = [m1u1 + m2u2] / (m1 + m2)

                                                      v = 0.88 m/s

Part D:

The velocity of ball after collision = The velocity of ball after collision

                                                 v = 0.88 m/s

We have, (1/2)m1v^2 = (1/2)kx^2

x = [m1v^2 / k]^(1/2) = 0.06 m

So the compression in the spring, x = 0.06 m

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