Question

A small 1.0 kg steel ball rolls west at 3.0 m/s collides with a large 3.0 kg ball at rest. After the collision, the small ball moves south at 2.0 m/s. What is the direction of the momentum (with respect to east) of the large ball after the collision

Answer 1

Answer:

The direction of the momentum of the large ball after the collision with respect to east is 146.58°.

Explanation:

Given that,

Mass of large ball = 3.0 kg

Mass of steel ball = 1.0 kg

Velocity = 3.0 kg

After collision,

Velocity = 2.0 m/s

Using conservation of momentum

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

$3.0\times0+1.0\times(3.0)(-i)=1.0\times2(-j)+3.0\times v_{2}$

$-3i+2j=3.0\times v_{2}$

$v_{2}=-i+0.66j$

The direction of the momentum

$tan\theta=\dfrac{0.66}{-1}$

$\theta=tan^{-1}\dfrac{0.66}{-1}$

$\theta=-33.42^{\circ}$

The direction of the momentum with respect to east

$\theta=180-33.42=146.58^{\circ}$

Hence, The direction of the momentum of the large ball after the collision with respect to east is 146.58°.