Question

The potential energy-displacement graph of a 0.5 kg ball moving along a frictionless track is shown in the Fig. At O, the velocity of the ball is 0 m s–1 and potential energy is 30 J. Calculate the velocity of the ball at P, Q and R.

Answer 1

Given,

Mass (m) = 0.5 kg

At O, velocity = 0 m s^–1 and potential energy = 30 J

So, kinetic energy at O = 0 J

Total mechanical energy = Kinetic energy + Potential energy = 0 + 30 = 30 J

The whole mechanical energy (30 J) is conserved at every location since the track is frictionless.

At any point, Kinetic energy = Total mechanical energy – Potential energy

At P (potential energy = 20 J):

Kinetic energy = 30 – 20 = 10 J

`1/2 "mv"^2 = 10`

`1/2 xx 0.5 xx v^2 = 10`

v² = 40

`v = sqrt(40) = 2sqrt(10) ≈ 6.3  "m s"^-1`

At Q (potential energy = 30 J):

Kinetic energy = 30 – 30 = 0 J

v = 0 m s^–1

The ball briefly stops at Q and then turns around.

At R (potential energy = 40 J):

Kinetic energy = 30 – 40 = –10 J

The ball cannot go to point R because kinetic energy cannot be negative. The ball turns back before it reaches R because the potential energy at R (40 J) is higher than the ball's total mechanical energy (30 J).

Hence, the velocity at P is about 6.3 m s^–1, at Q is 0 m s^–1, and the ball does not reach R.