I have seen many solutions for this question, but i do not understand how the fo

Question

I have seen many solutions for this question, but i do not understand how the formula used is derived. Please give a very thorough explanation. Thanks! A wagon is rolling forward on level ground. Friction is negligible. The person sitting in the wagon is holding a rock. The total mass of the wagon, rider, and rock is 95.5 kg. The mass of the rock is 0.310 kg. Initially the wagon is rolling forward at a speed of 0.540 m/s. Then the person throws the rock with a speed of 15.0 m/s. Both speeds are relative to the ground. (a) Find the speed of the wagon after the rock is thrown directly forward. (b) Find the speed of the wagon after the rock is thrown directly backward.

Explanation / Answer

a)Applying conservation of momentum(forward positive) For rock thrown forwards Wagon momentum before = wagon momentum after + rock momentum. 95.5kg x 0.54m/s = (95.5kg x v) + (0.31kg x 15m/s) 95.5V =46.82 V=0.49m/s For rock thrown backwards . Wagon momentum before = wagon momentum after - rock momentum 95.5 x 0.54 = (95.5 x v) - (0.31 x 15) 56.22 = 95.5v V=0.59m/s

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