Answer:
- The final velocity of the queen is (3/2) of the initial velocity of the striker. That is, (3V/2)
- The final velocity of the striker is (1/2) of the initial velocity of the striker. That is, (V/2)
Hence, the relative velocity of the queen with respect to the striker after collision
= (3V/2) - (V/2)
= V m/s.
Explanation:
This is a conservation of Momentum problem.
Momentum before collision = Momentum after collision.
The mass of the striker = M
Initial Velocity of the striker = V (+x-axis)
Let the final velocity of the striker be u
Mass of the queen = (M/3)
Initial velocity of the queen = 0 (since the queen was initially at rest)
Final velocity of the queen be v
Collision is elastic, So, momentum and kinetic energy are conserved.
Momentum before collision = (M)(V) + 0 = (MV) kgm/s
Momentum after collision = (M)(u) + (M/3)(v) = Mu + (Mv/3)
Momentum before collision = Momentum after collision.
MV = Mu + (Mv/3)
V = u + (v/3)
u = V - (v/3) (eqn 1)
Kinetic energy balance
Kinetic energy before collision = (1/2)(M)(V²) = (MV²/2)
Kinetic energy after collision = (1/2)(M)(u²) + (1/2)(M/3)(v²) = (Mu²/2) + (Mv²/6)
Kinetic energy before collision = Kinetic energy after collision
(MV²/2) = (Mu²/2) + (Mv²/6)
V² = u² + (v²/3) (eqn 2)
Recall eqn 1, u = V - (v/3); eqn 2 becomes
V² = [V - (v/3)]² + (v²/3)
V² = V² - (2Vv/3) + (v²/9) + (v²/3)
(4v²/9) = (2Vv/3)
v² = (2Vv/3) × (9/4)
v² = (3Vv/2)
v = (3V/2)
Hence, the final velocity of the queen is (3/2) of the initial velocity of the striker and is in the same direction.
The final velocity of the striker after collision
= u = V - (v/3) = V - (V/2) = (V/2)
The relative velocity of the queen withrespect to the striker after collision
= (velocity of queen after collision) - (velocity of striker after collision)
= v - u
= (3V/2) - (V/2) = V m/s.
Hope this Helps!!!!
