A 0.10kg dart is fired horizontally at the center of a 0.80kg vertically hanging

Question

A 0.10kg dart is fired horizontally at the center of a 0.80kg vertically hanging block, sticking to it. The system reaches a maximum height, h, of 0.20m above the collision point. Find:
a) the velocity of the dart – block system just after the collision

b) the velocity of the dart before collision

c) the energy lost during the collision

d) if the dart gains its speed due to a spring system that is initially compressed 0.05m from its equilibrium position, determine the spring constant.

e) If the dart didn’t stick to the block, but instead fell vertically after collision, would the initial velocity of the dart have to be greater, less than, or equal to the velocity that was found in part (b)? Justify.

Explanation / Answer

a) (m + M) g h = 1/2 (m + M) v^2 v = (2 g h)^1/2 speed of dart-block system by conservation of energy b) m V = (m + M) v conservation of momentum where V is speed of dart before collision and v speed found in (a) c) E(lost) = 1/2 m V^2 - 1/2 (M + m) v^2 d) If the dart fell vertically then m V = M v and V = M v / m which is smaller than the value found in (b) (the momentum of the dart does not now have to be included in the final momentum so the block will have more momentum by itself than the combination of the block and dart - also one can say that the dart gives all of its momentum to the block in part (d) but not in part (b)

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