Laws Of Motion Question 45

Question: A body of mass M at rest explodes into three pieces, two of which of mass M/4 each are thrown off in perpendicular directions with velocities of 3 m/s and 4 m/s respectively. The third piece will be thrown off with a velocity of [CPMT 1990]

Options:

A) 1.5 m/s

B) 2.0 m/s

C) 2.5 m/s

D) 3.0 m/s

Show Answer

Answer:

Correct Answer: C

Solution:

Momentum of one piece $ =\frac{M}{4}\times 3 $

Momentum of the other piece $ =\frac{M}{4}\times 4 $

Resultant momentum $ =\sqrt{\frac{9M^{2}}{16}+M^{2}}=\frac{5M}{4} $

The third piece should also have the same momentum.

Let its velocity be v, then $ \frac{5M}{4}=\frac{M}{2}\times v $

or $ v=\frac{5}{2}=2.5m/sec $



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