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Question: Answered & Verified by Expert
A $1.0 \mathrm{~kg}$ ball drops vertically onto the floor with a speed of $25 \mathrm{~m} / \mathrm{s}$. It rebounds with an initial speed of $10 \mathrm{~m} / \mathrm{s}$. The impulse action on the ball during contact will be
PhysicsCenter of Mass Momentum and CollisionJEE Main
Options:
  • A 15 N-s
  • B 25 N-s
  • C 35 N-s
  • D 45 N-s
Solution:
1191 Upvotes Verified Answer
The correct answer is: 35 N-s
$\begin{aligned} & \mathrm{m}=1.0 \mathrm{~kg}, \mathrm{v}_1=25 \mathrm{~m} / \mathrm{s}, \mathrm{v}_2=10 \mathrm{~m} / \mathrm{s} \\ & \text { Impulse }=\text { change in momentum } \\ & \mathrm{J}=\Delta \mathrm{P} \\ & \mathrm{J}=\mathrm{m}\left(\mathrm{v}_1-\mathrm{v}_2\right) \\ & \mathrm{J}=1.0 \times(25-(-10)) \\ & =1 \times(25)-1 \times(-10)=35 \mathrm{~N}-\mathrm{s} \\ & \therefore \mathrm{J}=35 \mathrm{~N}-\mathrm{s}\end{aligned}$

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