JEE MAIN - Physics (2021 - 18th March Evening Shift - No. 3)
An object of mass m1 collides with another object of mass m2, which is at rest. After the collision the objects move with equal speeds in opposite direction. The ratio of the masses m2 : m1 is :
1 : 1
3 : 1
2 : 1
1 : 2
Explanation
Before collision
_18th_March_Evening_Shift_en_3_3.png)
After collision,
_18th_March_Evening_Shift_en_3_4.png)
Applying momentum conservation,
m1v1 = m2v $$-$$ m1v ..... (1)
As collision is elastic,
$$ \therefore $$ e = 1
$$ \Rightarrow $$ $${{v - ( - v)} \over {0 - {v_1}}} = 1$$
$$ \Rightarrow $$ v1 = 2v ..... (2)
Put value of v1 in equation (1),
m1 (2v) = (m2 $$-$$ m1)v
$$ \Rightarrow $$ 2m1 = m2 $$-$$ m1
$$ \Rightarrow $$ 3m1 = m2 $$ \Rightarrow $$ $${{{m_2}} \over {{m_1}}} = {3 \over 1}$$
After collision,
Applying momentum conservation,
m1v1 = m2v $$-$$ m1v ..... (1)
As collision is elastic,
$$ \therefore $$ e = 1
$$ \Rightarrow $$ $${{v - ( - v)} \over {0 - {v_1}}} = 1$$
$$ \Rightarrow $$ v1 = 2v ..... (2)
Put value of v1 in equation (1),
m1 (2v) = (m2 $$-$$ m1)v
$$ \Rightarrow $$ 2m1 = m2 $$-$$ m1
$$ \Rightarrow $$ 3m1 = m2 $$ \Rightarrow $$ $${{{m_2}} \over {{m_1}}} = {3 \over 1}$$
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