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 $\big|\mathbf{m_1 \vec{v}_1'} + \mathbf{m_2 \vec{v}_2'}\big| - (\mathbf{m_1 \vec{v}_1} + \mathbf{m_2 \vec{v}_2})$ | is
$(a) Zero (b) (m_{1} + m_{2}) g t_{0} (c) 2 (m_{1} + m_{2}) g t_{0} (d) \frac{1}{2} (m_{1} + m_{2}) g t_{0}$
Text Solution
Verified by ExpertsC
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 $(m_{1} + m_{2}) g \times 2 t_{0}$
∴ ∴ |Change in momentum in this interval
$= |m_{1}\vec{v}_{1}^{\prime} + m_{2}\vec{v}_{2}^{\prime} - (m_{1}\vec{v}_{1} + m_{2}\vec{v}_{2})| = 2(m_{1} + m_{2})gt_{0}$
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems