Work Energy And Power Question 75
Question: Two particles of masses $ m_1 $ and $ m_2 $ in projectile motion have velocities $ {{\vec{v}}_1} $ and $ {{\vec{v}}_2} $ respectively at time t = 0. They collide at time $ t_0 $ . Their velocities become $ {{\vec{v}}_1}’ $ and $ {{\vec{v}}_2}’ $ at time $ 2t_0 $ while still moving in air. The value of $ |(m_1\overrightarrow{v_1}’+m_2\overrightarrow{v_2}’)-(m_1\overrightarrow{v_1}+m_2\overrightarrow{v_2}) $ | is [IIT-JEE Screening 2001]
Options:
A) Zero
B) $ (m_1+m_2)gt_0 $
C) $ 2(m_1+m_2)gt_0 $
D) $ \frac{1}{2}(m_1+m_2)gt_0 $
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Answer:
Correct Answer: C
Solution:
The momentum of the two-particle system, at t = 0 is $ {{\vec{P}} _{i}}=m_1{{\vec{v}}_1}+m_2{{\vec{v}}_2} $ Collision between the two does not affect the total momentum of the system.
A constant external force $ (m_1+m_2)g $ acts on the system. The impulse given by this force, in time t = 0 to $ t=2t_0 $ is $ (m_1+m_2)g\times 2t_0 $ \ |Change in momentum in this interval $ =|m_1\vec{v}{’_1}+m_2\vec{v}{’_2}-(m_1{{\vec{v}}_1}+m_2{{\vec{v}}_2})|=2(m_1+m_2)gt_0 $
BETA
