A 2.5 kg mass on a frictionless, horizontal surface is pushed against a horizont

Question

A 2.5 kg mass on a frictionless, horizontal surface is pushed against a horizontal spring with a spring constant of 2000 N/m, compressing the spring a distance of 35 cm. It is released from rest at t=0.

A) What is the shortest time at which the speed of the mass is 2.0 m/s?

B) How long does it take for the mass to reach the equilibrium point of the spring?

C) When it reaches the equilibrium point of the spring, it collides with a motionless 1.0 kg mass and sticks to it. What is the new amplitude of the motion?

D) How long after the collision in C) will it be before the spring it once again at its equilibrium length?

Explanation / Answer

Force in spring= -kx= 2000*.35=700N

Mass= 2.5kg

Accln= force/mass=700/2.5= 280 m/s2

Velocity= 2m/s

Time taken= 280/2= 140s

b) using energy conservation theorem

1/2mv2=1/2 kx2

Mv2=kx2

Velocity when the spring reach the equilibrium v2=2000*(.35)2/2.5=98

V=9.89m/s

Time taken= distance/velocity=.35/9.89= .03 sec

New amplitude of motion when it strike with a ball of 1kg=

M1V1=m1v1+m2v2

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