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Question

Particles of mass m and M moving with horizontal veocity u1=4 m/sec and u2=6 m/s as shown in figure. If m<<M then for one dimensional elastic collision, the velocity of lighter particle after collision will be

A
4 m/s opposite to its original direction.
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B
16 m/s along its original direction.
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C
12 m/s opposite to its original direction.
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D
16 m/s opposite to its original direction.
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Solution

The correct option is D 16 m/s opposite to its original direction.
Before collision

After collision


From the concept of collision we know
v1=(mMm+M)u1+2Mu2m+M
[As m<<M]
v1=u1+2u2 =(4+2×(6)) m/s =16 m/s
Here ve indicates ball will have final velocity of 16 m/s opposite to its original direction.

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