More practice questions

  1. Q4. A neutron makes a head-on elastic collision with a carbon nucleus (12 times its mass) at rest. What fraction of its kinetic energy does it keep?

    Show answerHide answer f1=(1−121+12)2=(1113)2≈0.716\begin{aligned} f_1 &= \left(\frac{1 - 12}{1 + 12}\right)^{2} = \left(\frac{11}{13}\right)^{2} \\ &\approx 0.716 \end{aligned}

    It keeps about 71.6% and gives about 28.4% to the carbon nucleus.

  2. Q5. A ball dropped from 1 m bounces back to 0.64 m. Find the coefficient of restitution between the ball and the floor (a term beyond NCERT, defined on this page).

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    The speed just before the bounce goes as 2g×1\sqrt{2g \times 1} and just after as 2g×0.64\sqrt{2g \times 0.64}, so e=0.64/1e = \sqrt{0.64/1}, that is 0.8.

  3. Q6. A 1 kg ball moving at 6 m s catches up with a 2 kg ball moving at 3 m s in the same direction, and they collide head-on elastically. Find their velocities afterwards.

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    Use the formulas for a head-on elastic collision with both bodies moving (given on this page, beyond NCERT), with v1i = 6 and v2i = 3 m s:

    v1f=(1−2)(6)+2(2)(3)1+2=−6+123=2\begin{aligned} v_{1f} &= \frac{(1 - 2)(6) + 2(2)(3)}{1 + 2} \\ &= \frac{-6 + 12}{3} = 2 \end{aligned} v2f=(2−1)(3)+2(1)(6)1+2=3+123=5\begin{aligned} v_{2f} &= \frac{(2 - 1)(3) + 2(1)(6)}{1 + 2} \\ &= \frac{3 + 12}{3} = 5 \end{aligned}

    So the 1 kg ball moves on at 2 m s and the 2 kg ball at 5 m s, both in the original direction.

    Check: momentum 6 + 6 = 12 before and 2 + 10 = 12 after; kinetic energy 18 + 9 = 27 J before and 2 + 25 = 27 J after.