Question

At \(t=0\) two particles \(A\) of mass \(3.4\,kg\) and \(B\) of mass \(2.5\,kg\) are moving along \(x\)-axis with initial velocities \(5\,m/s\) and \(10\,m/s\) respectively starting from \(x=0\). At \(t=5s\) position of \(A\) is \(x=104\,m\) and of \(B\) is \(x=137\,m\). Find ratio of momentum at \(t=10s\).

• A. \(2.17\)
• B. \(0.17\)
• C. \(3.17\)
• D. \(1.17\)

Answer 1

The Correct Option is D

Concept: Use kinematic equation: \[ s = ut + \frac{1}{2}at^2 \] Then velocity at time \(t\): \[ v = u + at \] Momentum: \[ p = mv \]
Step 1:Find acceleration of particle A. \[ 104 = 5\times5 + \frac{1}{2}a(25) \] \[ 104 = 25 + \frac{25a}{2} \] \[ 79 = \frac{25a}{2} \] \[ a = \frac{79\times2}{25} \] Velocity at \(t=10s\): \[ v_A = u + at \] \[ v_A = 5 + \frac{79\times2}{25}\times10 \] \[ v_A = \frac{341}{5} \]
Step 2:Find acceleration of particle B. \[ 137 = 10\times5 + \frac{1}{2}a(25) \] \[ 137 = 50 + \frac{25a}{2} \] \[ 87 = \frac{25a}{2} \] \[ a = \frac{87\times2}{25} \] Velocity at \(t=10s\): \[ v_B = u + at \] \[ v_B = 10 + \frac{87\times2}{25}\times10 \] \[ v_B = \frac{398}{5} \]
Step 3:Find ratio of momenta. \[ \frac{p_A}{p_B} = \frac{m_A v_A}{m_B v_B} \] \[ = \frac{3.4 \times \frac{341}{5}}{2.5 \times \frac{398}{5}} \] \[ = 1.165 \] \[ \boxed{1.17} \]