MHT CET · Physics · Center of Mass Momentum and Collision
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
- A \(1 \mathrm{~N}-\mathrm{s}\)
- B \(5 \mathrm{~N}-\mathrm{s}\)
- C \(2 \mathrm{~N}\) -s
- D \(3 \mathrm{~N}-\mathrm{s}\)
Answer & Solution
Correct Answer
(B) \(5 \mathrm{~N}-\mathrm{s}\)
Step-by-step Solution
Detailed explanation
Impulse \(=\) Change is momentum \(p_{i}-p_{f}\) direction of \(p_{f}\) and \(p_{i}\) apposite to each other.
\(\begin{aligned}\therefore & \text { Impulse }=m u-(-m v) \\&=m u+m v=m(u+v)=0.1(30+20) \\&=0.1 \times 50=5 \mathrm{~N}-\mathrm{s}\end{aligned}\)
\(\begin{aligned}\therefore & \text { Impulse }=m u-(-m v) \\&=m u+m v=m(u+v)=0.1(30+20) \\&=0.1 \times 50=5 \mathrm{~N}-\mathrm{s}\end{aligned}\)
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