Question

A bullet fired from gun with a velocity 30 m/s
at an angle of 60° with horizontal direction. At
the highest point of its path, the bullet
explodes into two parts with masses in the
ratio 1:3. The lighter mass comes to rest
immediately. Then the speed of the heavier
mass is



(A) 30 m/s
B) 20 m/s
(C) 10 m/s​

Answer 1

done....

hope this helps you....!

Answer 2

If a bullet fired from gun with a velocity 30 m/s  at an angle of 60° with horizontal direction and no external force is applied then its linear momentum will be conserved.

So, by the law of conservation of momentum

P(initial)=P(final)-----------------eq 1

Let the initial mass of the bullet is 4m

After it explodes  into two parts with masses in the  ratio 1:3 i.e 1m and 3m

from eq 1

MV = mv' + 3mv''                       ( since P=mv)

4m*15 = 1*0 + 3m*v''          ( here M= mass initially, V= velocity initially

                                ( v'= velocity of lighter mas ; v''=velocity of heavier mass)

v''=20 m/s