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

Step 1: Understanding the Concept:
Assuming constant acceleration for both particles, we can find their respective accelerations using the given displacement data.
Once accelerations are known, we calculate their velocities at the final time to find momentum.
Step 2: Key Formula or Approach:
Kinematic equation for position: $S = ut + \frac{1}{2}at^2$.
Kinematic equation for velocity: $v = u + at$.
Momentum: $P = mv$.
Step 3: Detailed Explanation:
For particle A ($m_A = 3.4 \text{ kg}, u_A = 5 \text{ m/s}, t = 5 \text{ s}, S_A = 104 \text{ m}$):
\[ 104 = 5(5) + \frac{1}{2}a_A(5^2) \implies 104 = 25 + 12.5 a_A \]
\[ 12.5 a_A = 79 \implies a_A = \frac{79 \times 2}{25} = 6.32 \text{ m/s}^2 \]
Velocity of A at $t = 10 \text{ s}$:
\[ v_A = u_A + a_A t = 5 + 6.32(10) = 5 + 63.2 = 68.2 \text{ m/s} \]
Momentum of A: $P_A = 3.4 \times 68.2 = 231.88 \text{ kg}\cdot\text{m/s}$.
For particle B ($m_B = 2.5 \text{ kg}, u_B = 10 \text{ m/s}, t = 5 \text{ s}, S_B = 137 \text{ m}$):
\[ 137 = 10(5) + \frac{1}{2}a_B(5^2) \implies 137 = 50 + 12.5 a_B \]
\[ 12.5 a_B = 87 \implies a_B = \frac{87 \times 2}{25} = 6.96 \text{ m/s}^2 \]
Velocity of B at $t = 10 \text{ s}$:
\[ v_B = u_B + a_B t = 10 + 6.96(10) = 10 + 69.6 = 79.6 \text{ m/s} \]
Momentum of B: $P_B = 2.5 \times 79.6 = 199 \text{ kg}\cdot\text{m/s}$.
Ratio of momentum:
\[ \frac{P_A}{P_B} = \frac{231.88}{199} \approx 1.165 \approx 1.17 \]
Step 4: Final Answer:
The ratio of momentum is 1.17.