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32 votes
32 votes
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?

User Yusuke Hakamaya
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1 Answer

19 votes
19 votes

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)* 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*11=(0.35+0.45)* v*\cos 32^(\circ) \\ \Rightarrow v=(0.35*11)/((0.35+0.45)*\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.

User BoxerBucks
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