A body of mass 2 kg moving with a speed of 4 m/s. makes an elastic collision with another body at rest and continues to move in the original direction but with one fourth of its initial peed. The speed of the two body centre of mass is ${x \over {10}}$ m/s. Then the value of x is ___________.
Answer (integer)
25
Solution
p<sub>i</sub> = p<sub>f</sub><br><br>2 $\times$ 4 = 2 $\times$ 1 + m<sub>2</sub> $\times$ v<sub>2</sub><br><br>m<sub>2</sub>v<sub>2</sub> = 6 ..... (i)<br><br>by coefficient of restitution<br><br>$1 = {{{v_2} - 1} \over 4} \Rightarrow {v_2} = 5$ m/s<br><br>by (i)<br><br>m<sub>2</sub> $\times$ 5 = 6<br><br>m<sub>2</sub> = 1.2 kg<br><br>${v_{cm}} = {{{m_1}{v_1} + {m_2}{v_2}} \over {{m_1} + {m_2}}}$<br><br>$${v_{cm}} = {{2 \times 1 + 1.2 \times 5} \over {2 + 1.2}} = {8 \over {3.2}} = {{25} \over {10}}$$<br><br>x = 25
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
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