Question

A loaded spring gun, initially at rest on a horizontal frictionless surface fires a marble of mass $m$ at an angle of elevation $\theta$. The mass of the gun is M, that of the marble is $m$ and the muzzle velocity of the marble is $v_0$. then velocity of the gun just after the firing is:

• A. $\frac{m{v}_0\cos \theta }{M+m}$
• B. $\frac{m{v}_0\cos \theta }{M}$
• C. $\frac{m{v}_0\cos 2\theta }{M+m}$
• D. $\frac{m{v}_0}{M}$

Answer 1

The correct option is A $\frac{m{v}_0\cos \theta }{M+m}$

Step 1:
Draw a labelled diagram.

Step 2:
Find the velocity of the gun just after the firing.
Formula used:
$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$
$\text{Muzzle velocity}=v_{m/g}=v_0$
Along $x$−direction;
$v_{m\left(x\right)}-v_{g\left(x\right)}=v_0\cos \theta$
By momentum conservation:
$(M+m)\left(0\right)=m\left(\right(v_0\cos \theta -v)-Mv$
$v=\frac{m{v}_0\cos \theta }{(M+m)}$
Final Answer: (c)