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WAEC Physics 2023 Paper

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Question : 40

Total: 50

A lead bullet of mass 0.05 kg is fired with a speed of 200 m/s into a lead block of mass 0.95 kg at rest. Given that the lead block moves after the impact, determine its kinetic energy.

150J
200J
50J
100J

Solution:

\; M_{\text{bullet}} = 0.05\ \text{kg},\ V_{\text{bullet}} = 200\ \text{m} / \text{s},\ M_{\text{block}} = 0.95\ \text{kg}\;\text{and}\; V_{\text{block}} = 0

\;\; \text{ for an inelastic collision }\;

\; M_{\text{bullet}} \times V_{\text{bullet}} + M_{\text{block}} \times V_{\text{block}} = ( M_{\text{bullet}} + M_{\text{block}} ) v

\; 0.05 \times 200 + 0.95 \times 0 = (0.05 + 0.95) v

\; 10 + 0 = v

\; v = 10\ \text{m} / \text{s}

K . E

of the block

= \frac{1}{2} mv^2

\; K . E = \frac{1}{2} \times (0.05+0.95) \times 10^2 \;\text{ (bullet now part of the block) }\;

\; K . E = \frac{1}{2} \times 1 \times 100

\; = 50\ \text{J}

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