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Question: A gun (mass = M) fires a bullet (mass=m) with speed\({v_r}\), relative to the barrel of the gun whic...

A gun (mass = M) fires a bullet (mass=m) with speedvr{v_r}, relative to the barrel of the gun which is inclined at an angle of 60∘60^\circ with horizontal. The gun is placed over a smooth horizontal surface. Find the recoil speed of the gun.
A) v=Mvr2(M+m)v = \dfrac{{M{v_r}}}{{2\left( {M + m} \right)}}.
B) v=mvr(M+m)v = \dfrac{{m{v_r}}}{{\left( {M + m} \right)}}.
C) v=mvr2(M+m)v = \dfrac{{m{v_r}}}{{\sqrt 2 \left( {M + m} \right)}}.
D) v=mvr2(M+m)v = \dfrac{{m{v_r}}}{{2\left( {M + m} \right)}}.

Explanation

Solution

The gun fires a bullet and due to that the gun fells recoil as the linear momentum should be conserved. As the bullet has speed at an angle of 60∘60^\circ and the gun will recoil in horizontal direction and therefore the horizontal component should be considered.

Formula used: The formula for the conservation of the linear momentum is given by,
⇒m1v1=m2v2\Rightarrow {m_1}{v_1} = {m_2}{v_2}
Where the mass of body 1 ism1{m_1}, the velocity of body 1 isv1{v_1}, the mass of body 2 is m2{m_2} and the velocity of body 2 is v2{v_2}.

Complete step by step solution:
It is given in the problem that a gun (mass = M) fires a bullet (mass=m) with speedvr{v_r}, relative to barrel of the gun which is inclined at an angle of 60∘60^\circ with horizontal the gun is placed over a smooth horizontal surface we need to find the recoil speed of the gun.
The momentum of the bullet is equal to,
⇒pb=mvrcos⁡60∘\Rightarrow {p_b} = m{v_r}\cos 60^\circ………eq. (1)
Let v be the velocity after recoil .The momentum of the gun and bullet after firing of the bullet is equal to,
⇒pr=(M+m)v\Rightarrow {p_r} = \left( {M + m} \right)v………eq. (2)
The linear conservation of momentum says that the momentum of two bodies before and after the interaction is equal.
As the linear conservation of momentum is conserved therefore equating equation (1) and (2) we get,
⇒(M+m)v=mvrcos⁡60∘\Rightarrow \left( {M + m} \right)v = m{v_r}\cos 60^\circ
⇒(M+m)v=mvr(12)\Rightarrow \left( {M + m} \right)v = m{v_r}\left( {\dfrac{1}{2}} \right)
⇒2(M+m)×v=m×vr\Rightarrow 2\left( {M + m} \right) \times v = m \times {v_r}.
⇒v=m×vr2(M+m)\Rightarrow v = \dfrac{{m \times {v_r}}}{{2\left( {M + m} \right)}}.
The velocity of the gum after recoil isv=m×vr2(M+m)v = \dfrac{{m \times {v_r}}}{{2\left( {M + m} \right)}}.

The correct answer for this problem is option D.

Note: The momentum is defined as the product of mass and velocity. The unit of the momentum is kg−mkg - m. It is advised to students to understand and remember the formula of the momentum and the conservation of linear momentum.