A 130 g block on a frictionless table is firmly attached to one end of a spring

Question

A 130 g block on a frictionless table is firmly attached to one end of a spring with k = 28 N/m . The other end of the spring is anchored to the wall. A 22 g ball is thrown horizontally toward the block with a speed of 5.5 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

m1 =22 g , u1 =5.5 m/s, m2 =130 g , k =28 N/m

Part A:

From conservation of momentum

m1u1+m2u2 = m1v1+m2v2

(22*5.5)+0 = (22*v1)+(130*v2) ...(1)

for elastic collision v2 -v1 = u1 -u2

v2 - v1 = 5.5 ...(2)

from (1) and (2) we get

velocity of ball after collision v1 = - 3.9 m/s

v2 = 1.6 m/s

Part B:

From conservation of energy

(1/2)m2v2^2 = (1/2)kx^2

0.13*1.6*1.6 = 28*x^2

x = 0.109 m

Part C:

From conservation of momentum

m1u1+m2u2 = (m1+m2)v

(22*5.5)+0 = (22+130)v

velocity of ball after collision v = 0.796 m/s

Part D:

From conservation of energy

(1/2)m2v2^2 = (1/2)kx^2

0.13*0.796*0.796 = 28*x^2

x = 0.054 m

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