2. Solved Papers
  3. AMU Medical

AMU Medical AMU Solved Paper-2000

  • question_answer
    A 1.0 kg ball falls vertically on to the floor with a speed of 25 m/s. It rebounds with an initial speed of 10 m/s. Find the restitution coefficient and the impulse acting on the ball during contact, if the ball is in contact for 0.02 s. What is average exerted force on the floor ?

    A)  0.1, 55 kg m/s, 1550 N

    B)  0.2, 45 kg m/s 1650 N

    C)  0.3, 30 kg m/s, 1550 N

    D)  0.4, 35 kg m/s, 1750 N

    Correct Answer: D

    Solution :

                     Impulse = change in momentum             ... (i) Also, Impulse = force (F)                                 \[\times \]time duration (At)      ... (ii)                 \[\Delta p=m\Delta v=m\,(u-v)\]                                 \[=1\,(25-(-10)=35\,kg-m/s\]                 \[\therefore \]  Force \[=\frac{Impulse}{time\text{ }\left( \Delta t \right)}\]                 \[=\frac{35}{0.02}=1750\,N\]. Also, coefficient of restitution                 \[(e)=\frac{velocity\text{ }of\text{ }rebound}{velocity\text{ }of\text{ }fall}\]                 \[=\frac{10}{25}=0.4\]


You need to login to perform this action.
You will be redirected in 3 sec spinner