First slide
Second law
Question

A person of mass 72 kg sitting on ice pushes a block of mass of 30 kg on ice horizontally with a speed of 12 ms-1. The coefficient of friction between the man and ice and between block and ice in 0.02. If g =10 ms-2, the distances between man and the block, when they come to rest is

Moderate
Solution

From LCLM
m1v1 = m2v2 ⇒ 30 x 12 = 72v2
{v_2} = \frac{{\mathop {\cancel{{30}}}\limits^5 \times \cancel{{12}}}}{{\mathop {\cancel{{72}}}\limits_{\cancel{6}} }} = 5m/s
S = {S_1} + {S_2} = \frac{{v_1^2}}{{2\mu g}} + \frac{{v_2^2}}{{2\mu g}} = \frac{1}{{2\mu g}}[v_1^2 + v_2^2]
= \frac{1}{{2 \times 0.02 \times 10}}[{12^2} + {5^2}]
= \frac{1}{{0.4}}[144 + 25] = \frac{{169}}{{0.4}} \Rightarrow S = \frac{{\mathop {\cancel{{169}}}\limits^{42.25} \times 10}}{{\cancel{4}}} = 422.5m

   
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