JEE MAIN - Physics (2021 - 25th July Morning Shift - No. 17)
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
25
Explanation
pi = pf
2 $$\times$$ 4 = 2 $$\times$$ 1 + m2 $$\times$$ v2
m2v2 = 6 ..... (i)
by coefficient of restitution
$$1 = {{{v_2} - 1} \over 4} \Rightarrow {v_2} = 5$$ m/s
by (i)
m2 $$\times$$ 5 = 6
m2 = 1.2 kg
$${v_{cm}} = {{{m_1}{v_1} + {m_2}{v_2}} \over {{m_1} + {m_2}}}$$
$${v_{cm}} = {{2 \times 1 + 1.2 \times 5} \over {2 + 1.2}} = {8 \over {3.2}} = {{25} \over {10}}$$
x = 25
2 $$\times$$ 4 = 2 $$\times$$ 1 + m2 $$\times$$ v2
m2v2 = 6 ..... (i)
by coefficient of restitution
$$1 = {{{v_2} - 1} \over 4} \Rightarrow {v_2} = 5$$ m/s
by (i)
m2 $$\times$$ 5 = 6
m2 = 1.2 kg
$${v_{cm}} = {{{m_1}{v_1} + {m_2}{v_2}} \over {{m_1} + {m_2}}}$$
$${v_{cm}} = {{2 \times 1 + 1.2 \times 5} \over {2 + 1.2}} = {8 \over {3.2}} = {{25} \over {10}}$$
x = 25
Comments (0)
