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 $
