Work Energy And Power Question 84
Question: A body of mass 5 kg explodes at rest into three fragments with masses in the ratio 1 : 1 : 3. The fragments with equal masses fly in mutually perpendicular directions with speeds of 21 m/s. The velocity of the heaviest fragment will be [CBSE PMT 1991]
Options:
A) 11.5 m/s
B) 14.0 m/s
C) 7.0 m/s
D) 9.89 m/s
Show Answer
Answer:
Correct Answer: D
Solution:
$ P _{x}=m\times v _{x}=1\times 21=21\ kg\ m/s $
$ P _{y}=m\times v _{y}=1\times 21=21\ kg\ m/s $
Resultant = $ \sqrt{P_x^{2}+P_y^{2}}=21\sqrt{2} $ kg m/s
The momentum of heavier fragment should be numerically equal to resultant of $ {{\vec{P}} _{x}} $
and $ {{\vec{P}} _{y}} $ . $ 3\times v=\sqrt{P_x^{2}+P_y^{2}}=21\sqrt{2} $
$ v=7\sqrt{2} $ = 9.89 m/s
