A 17.1 kg package is released from rest down a 34.0° chute. When the package rea

Question

A 17.1 kg package is released from rest down a 34.0° chute. When the package reaches the bottom it undergoes a perfectly elastic collision with a 34.2 kg stationary package, giving this second package a speed of 7.22 m/s. Assuming the chute is frictionless, what height must the first package have been released from?

Dana has set up a target for her toy spring gun. The gun requires 9.60 g pellets as ammunition that travel 12.3 m/s when shot. If the spring inside has maximum compression of 0.0900 m, what is the spring constant for the spring in the toy gun?

Explanation / Answer

1]

The collision at the bottom is perfectly elastic and the 2nd package is moving with a speed of 7.22 m/s,

so, using conservation of momentum,

mv + 0 = mV + MV '

=> 17.1v = 17.1V + 34.2(7.22)

=> v - V = 14.44 m/s

=> V = v - 14.44

and the kinetic energy is also conserved. So,

mv2 + 0 = mV2 + MV ' 2

=> 17.1v2 = 17.1V2 + 34.2(7.22)2

=> v2 - V2 = 104.257

=> v2 - [v - 14.44]2 = 104.257

=> 2(14.44)v - 208.5136 = 104.257

=> v = 10.83 m/s

this is the speed of the first package just before collision.

so, using conservation of energy

mgh = (1/2)mv2

=> h = (v2/2g) = 10.832/(2 x 9.8) = 5.984 m

this is the height required for the 1st package.

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