Momentum

A ball B 1 B_1 of mass 2 M '2 M' moving along x x -axis with uniform velocity 2 V '2 V' .Another ball B 2 B_2 of mass M 'M' with velocity 3 V '3 V' moving in same direction .Let p 1 p_1 and p 2 p_2 be the momentum of ball B 1 B_1 and B 2 B_2 respectively.Then the value of p 1 p 2 \frac{p_1}{p_2} is:

2 2 3 3 4 3 \frac{4}{3} 3 4 \frac{3}{4}

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1 solution

Akshay Sharma
Feb 11, 2016

We know that p = m v p=m v .For B 1 B_1 , momentum p 1 p_1 is p 1 = ( 2 M ) ( 2 V ) p_1 =(2 M) (2 V) = > => p 1 = 4 M V p_1=4 M V .For B 2 B_2 , momentum p 2 = 3 M V p_2=3 M V .Hence p 1 p 2 = 4 3 \frac{p_1}{p_2} =\frac{4}{3} .

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