Question

An object initially at rest explodes into three fragments A, B and C. The momentum of A is p $\hat{i}$ and that of B is $\sqrt{3}p\hat{j}$ where p is a +ve number. The momentum of C will be :

• A. $(1+\sqrt{3})p$ in a direction making angle 120${}^o$ with that of A.
• B. $(1+\sqrt{3})p$ in a direction making angle 150${}^o$ with that of B.
• C. 2p in a direction making angle 150${}^o$ with that of A.
• D. 2p in a direction making angle 150${}^o$ with that of B.

Answer 1

Answer: (D)

2p in a direction making angle 150${}^o$ with that of B.

Let the fragments $C$ moves with momentum $P_3$ making an angle $\theta$ with negative x axis.

Initial momentum of the bomb is Zero i.e $P_i=0$

Applying conservation of momentum in x direction :

$0=p-P_{3}cos\theta$ $⟹P_{3}cos\theta =p$ ...............(1)

Applying conservation of momentum in y direction :

$0=\sqrt{3}p-P_{3}sin\theta$ $⟹P_{3}sin\theta =\sqrt{3}p$ ...............(2)

Squaring and adding (1) and (2), $P_3^2=p^2+3p^2$

$⟹P_3=2p$

Now dividing (1) form (2) we get, $tan\theta =\sqrt{3}$ $⟹\theta ={60}^o$

Angle made by $P_3$ with $P_2$ $\varphi ={90}^o+\theta ={90}^o+{60}^o={150}^o$