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Question

Physics Question on Oscillations

A spring of natural length ll and spring constant 50N/m50\, N/m is kept on a horizontal frictionless table with one end attached to a rigid support. First the spring was compressed by 10cm10\, cm and then released to hit a ball of mass 20g20\, g kept at a distance l from the rigid support. if after hitting the ball, the natural length of the spring is restored, what is the speed with which the ball moved? (Ignore the air resistance)

A

5 m/s

B

7 m/s

C

25 m/s

D

50 m/s

Answer

5 m/s

Explanation

Solution

A spring of spring constant 50N/m50\, N / m is attached to a rigid wall. A 20g20\, g ball is placed on the horizontal table as shown in figure If spring is compressed then potential energy U=12kx2U=\frac{1}{2} k x^{2} When spring hit the ball, it transfer all its potential energy to kinetic energy of ball 12kx2=KEball\frac{1}{2} k x^{2}= KE _{ ball } 12kx2=12mv2\Rightarrow \frac{1}{2} k x^{2} =\frac{1}{2} m v^{2} v=kx2m\Rightarrow v=\sqrt{\frac{k x^{2}}{m}} Given, x=20cm,m=20gx=20\,cm , m=20\, g and v=50N/mv=50\, N / m v=50×(20×102)220×103\Rightarrow v =\sqrt{\frac{50 \times\left(20 \times 10^{-2}\right)^{2}}{20 \times 10^{-3}}} =25=5m/s=\sqrt{25}=5\, m / s