Spring mass collision A 3.0kg mass slides on a frictionless horizontal surface,

Question

Spring mass collision A 3.0kg mass slides on a frictionless horizontal surface, initially moving to the left at 50m/s. It collides with a spring, compressing it, and momentarily coming to rest. Finally, the mass moves to the right at 40m/s. Assume the block remains in contact with the spring for 0.02s. a.) What is the impulse of the spring on the block? Remember impulse is a vector. b.) What is the average force of the spring on the block? c.) If the spring has a stiffness constant of k = 50,000N/m, how much is the spring compressed during the collision?

Explanation / Answer

Given that

mass m=3 kg

initial velocity u=-50 m/s

final velocity v=40 m/s

time t=0.02 s

step;2

now we find the impule of mass

impulse J=m[v-u]=3[40+50]=270 kg.m/s

step;3

now we find the average forece of spring block moves

the average force Favg=J/t=270/0.02=13500 N

now we find the spring compression during collision

Favg=kx

13500=50000*x

distance x=0.27 m

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