Question

A .35kg hoover ball is sliding across the floor at 11m/s due East when it hits a hairball wrapped in gum(.045kg) sitting on the floor. The hoover ball and hair/gumball stick together and end up going 32 degrees South of East after the collision. What is the resultant velocity of the hoover/hair/gumball?

Answer 1

Given,

The mass of the hoover ball, m₁=0.35 kg

The velocity of the hoover ball before the collision, u=11 m/s

The mass of the hair/gumball, m₂=0.45 kg

The direction in which the hoover/hair/gumball goes after the collision, θ=32°

Let us assume that the eastward direction represents the positive x-axis and the northward direction represents the positive y-axis.

According to the law of conservation of momentum, the momentum is conserved in x and y directions separately and simultaneously.

Thus,

$m_1u=(m_1+m_2)\times v\cos \theta$

Where v is the velocity of the hoover/hair/gumball after the collision.

On substituting the known values,

$\begin{gathered} 0.35\times11=(0.35+0.45)\times v\times\cos 32^{\circ} \\ \Rightarrow v=\frac{0.35\times11}{(0.35+0.45)\times\cos 32^{\circ}} \\ =5.67\text{ m/s} \end{gathered}$

Thus the resultant velocity of the hoover/hair/gumball after the collision is 5.67 m/s.