Question

As per the given figure, a small ball $P$ slides down the quadrant of a circle and hits the other ball $Q$ of equal mass which is initially at rest Neglecting the effect of friction and assume the collision to be elastic, the velocity of ball $Q$ after collision will be :$\left( g =10 \,m / s ^2\right)$

• A. $4\, m / s$
• B. 0
• C. $0.25 \,m / s$
• D. $2\, m / s$

Answer 1

The Correct Option is D

To solve the problem of finding the velocity of ball \(Q\) after an elastic collision, we apply the principles of physics, particularly conservation of energy and conservation of momentum. 

1. Initial Energy Calculation:
   • Ball \(P\) starts sliding from the top of the quadrant. At the top, it only has potential energy.
   • The potential energy at the top is given by: \(PE = m \cdot g \cdot h\)
   • Since \(h = 20 \, \text{cm} = 0.2 \, \text{m}\), the energy is: \(PE = m \cdot 10 \cdot 0.2 = 2m\) (in joules)
2. Energy at the Bottom:
   • At the bottom of the quadrant, the potential energy is converted into kinetic energy (neglecting friction).
   • Kinetic energy at the bottom of the arc: \(KE = \frac{1}{2} m v^2 = 2m\)
   • Solving for \(v\): 
     \(\frac{1}{2} m v^2 = 2m\)
     \(v^2 = 4 \cdot 2\)
     \(v = 2 \,\text{m/s}\)
3. Conservation of Momentum:
   • Initially, ball \(Q\) is at rest, so initial momentum: \(m \cdot 2\text{ (velocity of P)} + m \cdot 0\text{ (velocity of Q)} = 2m\)
   • For a perfectly elastic collision, the two balls will exchange velocities since they are of equal mass.
   • Thus, after the collision, ball \(Q\) will have the velocity \(2 \, \text{m/s}\).

Therefore, the velocity of ball \(Q\) after collision will be \(2\, \text{m/s}\).