Question

1Q) a cannon of mass m1 = 12000kg fires a shell of mass m2 = 300kg in a horizontal direction with a velocity v2 = 400 m/s . find the velocity of the cannon after it is shot .

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Velocity\:of\:cannon=10\:m/s\:\:(Backward)}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Mass \: of \: cannon( m_{1} ) = 12000 \: kg \\ \\ \tt: \implies Mass \: of \:shell( m_{2} ) = 300 \: kg \\ \\ \tt: \implies Velocity \: of \: shell(v_{1}) = 400 \: m/s \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Velocity \: of \: cannon \: after \: it \: is \: shot( v_{2}) = ?$

• According to given question :

$\tt\circ\:Initial\:velocity\:of\:cannon(v)=0\:m/s \\\\\bold{As \: we \: know \: that} \\ \tt: \implies m_{1} v = m_{1} v_{ 1} + m_{2} v_{2} \\ \\ \tt: \implies 12000 \times 0= 12000 \times v_{1} + 300 \times 400 \\ \\ \tt: \implies 0 = 12000 \times v_{1} + 120000 \\ \\ \tt: \implies - 120000 = 12000 \times v_{1} \\ \\ \tt: \implies v_{1} = \frac{ - 120000}{12000} \\ \\ \green{\tt: \implies v_{1} = - 10 \: m/s} \\ \\ \green{\tt \therefore Velocity \: of \: cannon \: after \: shot \: is \: 10 \: m/s \: backward}$