Question

A box of mass 8 kg is moving at 9 m/s and collides with a 16 kg stationaryball.After the collision, the box continues in the same direction with avelocity of 3 m/s. Neglecting friction, the velocity of the ball after them/s. *collision is -

Answer 1

According to law of conservation of momentum we have,

$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}_{}_{}$

Where,

$m_1=8kg,m_2=16\operatorname{kg},v_{1i}=9\text{ m/s, }v_{2i}=0m/s,v_{1f}=3\text{ m/s}$

Substituting these values in the above equation,

$8\times9+16\times0=8\times3+16\times v_{2f}$$v_{2f}$

V2f is what we need to find out

Simplifying the above equation,

$72=24+16\times v_{2f}$

which implies

$v_{2f}=\frac{72-24}{16}=\frac{48}{16}$

Which gives us

$v_{2f}=3$

i.e. The velocity of the ball after the collision is 3 m/s