2. A 10-grams stone is shot horizontally at a 2.0-kg wooden block hooked to a re

Question

2. A 10-grams stone is shot horizontally at a 2.0-kg wooden block hooked to a relaxed 200 N/m spring (the other end of the spring is fixed against a wall, the spring and the block sitting on a horizontal frictionless table). It is observed that the stone bounces back at 10 m/s and that the block compresses the spring by 4.0 cm. You are dealing here with an inelastic collision (when total momentum is conserved) followed by a spring compression (when mechanical energy s conserved), a) What was the speed of the block right after the stone bounces off the block? b) What was the speed of the stone just before collision?

Explanation / Answer

A) Apply, 0.5*k*x^2   = 0.5*M*v^2

==> v = x*sqrt(k/m)

= 0.04*sqrt(200/2)

= 0.4 m/s

B) Let u is the initial speed of the stone before th collsion.

Apply momentum conservation.

m*u = m*v + M*V

0.01*u = 0.01*(-10) + 2*0.4

0.01*u = -0.1 + 0.8

u = 0.7/0.01

= 70 m/s

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