Question

A 0.2 kg block compresses a spring of spring constant 1900 N/m by 0.18 m. After being released from rest, the block slides along a smooth, horizontal and frictionless surface before colliding elastically with a 1.4 kg block which is at rest. (Assume the initial direction of motion of the sliding block before the collision is positive.)

A: What is the velocity of the 0.2 kg block just before striking the 1.4 kg block?

B: What is the velocity of the 1.4 kg block after the collision?

C: What is the velocity of the 0.2 kg block after the collision?

D: If the 0.2 kg block comes back into contact with the spring, what is the maximum compression of the spring?

Answer 1

part A:

for intial velocity, use Elastic pot energy = KE

0.5 kx^2 = 0.5 mv^2

v^2 = kx^2/m

v = x* sqrt(k/m)

v = 0.18* sqrt(1900/0.2)

v = 17.54 m/s

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b)here m1 = 0.2 kg, u1 = 17.54 m/s

m2 = 1.4 kg, u2 = 0

let v1 and v2 are the velocities of m1 and m2 after the collsion.

v2 = 2*m1*u1/(m1+m2)

v2 = 2*0.2*17.54/(0.2+1.4)

v2 = 4.385 m/s
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c) v1 = (m1-m2)*u1/(m1+m2)

v1 = (0.2 - 1.4)*17.54/(0.2+1.4)

v1 = -13.155m/s

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d) Again apply energy coservation

0.5*k*x^2 = 0.5*m*v^2

x = v*sqrt(m/k)

x = 13.155*sqrt(0.2/1900)

x 13.49 cm