Test Index

JAMB Physics 2021 Paper

Question : 21

Total: 40

^{-1}

Solution:

Concept:

After impact, the bullet and block move together, so kinetic energy is found using the common velocity from conservation of momentum.

Explanation:

Total mass after impact is

m_1 + m_2 = 0.05 + 0.95 = 1.00 \, \mathrm{kg}

Using conservation of momentum:

m_1 u_1 = (m_1 + m_2)v

0.05 \times 200 = 1.00 \times v

v = \frac{10}{1.00} = 10 \, \mathrm{m\,s^{-1}}

Final kinetic energy:

K.E = \frac{1}{2}(m_1 + m_2)v^2

K.E = \frac{1}{2}(1.00)(10)^2 = \frac{1}{2} \times 100 = 50 \, \mathrm{J}

Answer:

The final kinetic energy after impact is

50 \, \mathrm{J}

, which is option A.

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