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The potential energy-displacement graph of a 0.5 kg ball moving along a frictionless track is shown in Fig. 7.39. 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.
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Frictionless, so total mechanical energy = value at O = 30 J (all potential there). At any point, \(KE = 30 - PE\) and \(v = \sqrt{\dfrac{2\,KE}{m}} = \sqrt{4\,KE}\) (since m = 0.5 kg).
At P (PE = 20 J): \(KE = 10\ \text{J} \Rightarrow v = \sqrt{40} \approx 6.3\ \text{m s}^{-1}\).
At Q (PE = 30 J): \(KE = 0 \Rightarrow v = 0\ \text{m s}^{-1}\).
At R (PE = 40 J): this needs 40 J but only 30 J is available, so the ball cannot reach R — it turns back before R. (Read the exact PE values from your copy of Fig. 7.39.)
