The recoil velocity of a 4.0 kg rifle that shoots a 0.050 kg bullet at a speed of 280 text ms-1 is:

Question

The recoil velocity of a 4.0 kg rifle that shoots a 0.050 kg bullet at a speed of 280 text ms-1 is: (A)  +3.5 text ms-1  (B)  -3.5 text ms-1  (C)  -√(3.5) text ms-1  (D)  +√(3.5) text ms-1

Tardigrade Admin · Accepted Answer

m1u1+m2u2=0 ⇒ u1=-(m2u2/m1) =(0.050× 280/4)=-3.5 ms-1

Q. The recoil velocity of a $4.0 k g$ rifle that shoots a $0.050 k g$ bullet at a speed of $280 m s^{- 1}$ is:

$+ 3.5 m s^{- 1}$

$- 3.5 m s^{- 1}$

$- (3.5) m s^{- 1}$

$+ (3.5) m s^{- 1}$

$+7 m s^{- 1}$

$m_{1} u_{1} + m_{2} u_{2} = 0$
$\Rightarrow$ $u_{1} = - \frac{m _{2} u _{2}}{m _{1}}$
$= \frac{0.050 \times 280}{4} = - 3.5 m s^{- 1}$

Q. The recoil velocity of a 4.0kg rifle that shoots a 0.050kg bullet at a speed of 280ms−1 is:

+3.5ms−1

−3.5ms−1

−(3.5)​ms−1

+(3.5)​ms−1

+7 ms−1

m1​u1​+m2​u2​=0 ⇒ u1​=−m1​m2​u2​​ =40.050×280​=−3.5ms−1