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Question

A block of mass 1 kg travelling in a straight line with a velocity of 10 ms1 collides with and sticks to a stationary wooden block of mass 5 kg.
Then they both move off together in the same straight line. Calculate the total momentum just before the impact, just after the impact and the velocity of the combined object.

A
10 kg m s1, 20 kg m s1, 1.67 ms1
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B
10 kg m s1, 10 kg m s1, 2 ms1
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C
10 kg m s1, 10 kg m s1, 1.67 ms1
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D
5 kg m s1, 10 kg m s1, 2 ms1
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Solution

The correct option is C 10 kg m s1, 10 kg m s1, 1.67 ms1
Given,
Mass of first block, m1=1 kg
Mass of second block, m2=5 kg
Initial velocity of first block, u1=10 m/s
Initial velocity of second block, u2=0 m/s

The momentum before collision is:
pi=m1u1+m2u2
pi=1×10+5×0
pi=10 kg m/s

Let the final velocity of the blocks be v.
Final momentum will be:
pf=(m1+m2)v
pf=(1+5)v=6v

From the law of conservation of momentum, momentum before collision should be equal to momentum after the collision as no external force was acting on the bodies. So,
pf=pi
6v=10
v=1.67 m/s

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