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A ball of mass $0.1 \mathrm{~kg}$ strikes a wall normally with a speed of $30 \mathrm{~ms}^{-1}$ and rebounds with a speed of $20 \mathrm{~ms}^{-1}$. The impulse of the force exerted by the wall on the ball is
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Verified Answer
The correct answer is:
$5 \mathrm{~N}-\mathrm{s}$
Impulse $=$ Change is momentum $p_{i}-p_{f}$ direction of $p_{f}$ and $p_{i}$ apposite to each other.
$$
\begin{aligned}
\therefore \quad & \text { Impulse }=m u-(-m v) \\
&=m u+m v=m(u+v)=0.1(30+20) \\
&=0.1 \times 50=5 \mathrm{~N}-\mathrm{s}
\end{aligned}
$$
$$
\begin{aligned}
\therefore \quad & \text { Impulse }=m u-(-m v) \\
&=m u+m v=m(u+v)=0.1(30+20) \\
&=0.1 \times 50=5 \mathrm{~N}-\mathrm{s}
\end{aligned}
$$
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