A body of mass $4 \mathrm{~kg}$ is moving with momentum of $8 \mathrm{~kg} \mathrm{~ms}^{-1}$. A force of $0…

A body of mass $4 \mathrm{~kg}$ is moving with momentum of $8 \mathrm{~kg} \mathrm{~ms}^{-1}$. A force of $0.2 \mathrm{~N}$ acts on it in the direction of motion of the body for 10 seconds. The increase in kinetic energy in joules is
  1. $10$
  2. $8.5$
  3. $4.5$
  4. $4$

Solution

$ \begin{aligned} m & =4 \mathrm{~kg}, p=8 \mathrm{~kg} \mathrm{~m} / \mathrm{s} \\ F & =0.2 \mathrm{~N}, t=10 \mathrm{sec} \end{aligned} $ From Newton's IInd law Force $=$ rate of change of momentum $ \begin{aligned} & F=\frac{p_2-p_1}{t} \\ & 0.2=\frac{m v-8}{10} \\ & 2=4 \times v-8 \\ & 4 v=10 \\ & \Rightarrow \quad v=\frac{5}{2} \mathrm{~m} / \mathrm{s} \\ & \therefore \quad \text { Final momentum } p_2=m v=4 \times \frac{5}{2} \\ & =10 \mathrm{~kg} \mathrm{~m} / \mathrm{s} \\ & \end{aligned} $

Asked in: AP EAMCET 2002

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