Question

A $3000\text{kg}$ cannon recoils with a velocity of $1{\text{m s}}^{-1}$, when a $15\text{kg}$ cannon ball is fired from it. If the ball is aimed a target $400\text{m}$ away, calculate the force (in $\text{N}$) that the ball exerts on the target, when it hits it.

• A. 750

Answer 1

The correct option is A 750
Given
The mass of the cannon ball, ${\text{m}}_1=15\text{kg}$
The mass of the cannon, ${\text{m}}_2=3000\text{kg}$
Distance to the target, $\text{d}=400\text{m}$
The recoil velocity of the cannon, ${\text{v}}_2=1{\text{m s}}^{-1}$

If the velocity of the cannon ball is ${\text{v}}_1$, then according to the law of conservation of momentum,
${\text{m}}_1{\text{v}}_1+{\text{m}}_2{\text{v}}_2=0$, as both the cannon and the cannon ball are initially at rest.

${\text{v}}_1=\frac{-{\text{m}}_2{\text{v}}_2}{{\text{m}}_1}=\frac{3000}{15}=200{\text{m s}}^{-1}$

According to the third equation of motion,
${\text{v}}^2-{\text{u}}^2=2\text{as}$

$\text{v}=200{\text{m s}}^{-1};\text{u}=0;\text{s}=\text{d}=400\text{m}$

The acceleration of the cannon ball,
$\text{a}=\frac{{\text{v}}^2}{2\text{s}}=\frac{200\times 200}{2\times 400}=50{\text{m s}}^{-2}$

The force exerted by the cannon ball on the target is given by Newton's second law of motion as: $\text{F =}{\text{m}}_1\text{a}$

$⟹\text{F}=15\text{kg}\times 50{\text{m s}}^{-2}$

$⟹\text{F}=750\text{N}$