Q127 / 160 · TS EAMCET 2024 (09 May Shift 2) · Physics

A man of mass 80 kg goes to the market on a scooter of mass 100 kg with certain speed. On application of brakes, the stopping distance is $S_1$. The man returns home on the same scooter, with the same speed with a 60 kg bag of rice. If $S_2$ is the new stopping distance when the brakes are applied with the same force, then

  • A
    $7 S_1=4 S_2$
  • B
    $2 S_1=S_2$
  • C
    $3 S_1=4 S_2$
  • Correct answer
    D
    $4 S_1=3 S_2$

Solution

When stopping distance is $\mathrm{s}_1$ $v^2-u^2=2 a_1 s_1 \Rightarrow 0-u^2=2 \frac{F}{100+80} s_2 \qquad ...\mathrm{(i)}$ When stopping distance is $\mathrm{s}_2$ $v^2-u^2=2 a_2 s_2 \Rightarrow 0-u^2=2 \frac{F}{100+80+60} \times s_2 \qquad ...\mathrm{(ii)}$ From eq. (i) and (ii) $2 \frac{\mathrm{~F}}{180} \times \mathrm{s}_1=2 \frac{\mathrm{~F}}{240} \times \mathrm{s}_2 \therefore 4 \mathrm{~s}_1=3 \mathrm{~s}_2$

Hence, the correct answer is option: 'D'.

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