Question

A man (mass = 50 kg) and his son (mass = 20 kg) are standing on a frictionless surface facing each other. The man pushes his son so that he starts moving at a speed of 0.70 ms⁻¹ with respect to the man. The speed of the man with respect to the surface is :
(A) 0.47 ms⁻¹
(B) 0.28 ms⁻¹
(C) 0.14 ms⁻¹
(D) 0.20 ms⁻¹

Answer 1

Answer:

(B)

Explanation:

the answer is B.

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Answer 2

$\displaystyle\large\boxed {\sf {(D)\quad\!0.20\ m\ s^{-1}}}$

After pushing the son moves with a velocity $\displaystyle\bf {v_s}$ wrt ground.

Since the frictionless surface is not fixed, the surface also moves due to the motion of the son, but in the opposite direction, and hence the man also moves. Let his velocity due to this motion be $\displaystyle\bf {-v_m}$ wrt ground. Negative direction is because man moves opposite to the son.

Given,

$\displaystyle\longrightarrow\bf {(v_s)-(-v_m)}\ =\ \sf{0.7}$

$\displaystyle\longrightarrow\bf {v_s+v_m}\ =\ \sf{0.7}$

$\displaystyle\longrightarrow\bf{v_s}\ =\ \sf{0.7\ -\ }\bf{v_m}$

No external force is acting on the system, hence linear momentum is conserved, i.e.,

$\displaystyle\longrightarrow\sf{50}(-\bf{v_m})+\sf{20}\bf{v_s}\ =\ \sf{0}$

since both were in rest initially.

Then,

$\displaystyle\longrightarrow\sf{5}\bf{v_m}=\sf{2}\bf{v_s}$

$\displaystyle\longrightarrow\sf{5}\bf{v_m}=\sf{2(0.7\ -\ }\bf{v_m})$

$\displaystyle\longrightarrow\sf{5}\bf{v_m}=\sf{1.4\ -\ }\bf{2v_m}$

$\displaystyle\longrightarrow\sf{7}\bf{v_m}=\sf{1.4}$

$\displaystyle\longrightarrow\underline {\underline {\bf{v_m}\ =\ \sf{0.20\ m\ s^{-1}}}}$