Question

Two identical masses A and B are on same vertical line. A is released from rest while B is simultaneously projected upwards with a velocity of 50 m/s. Find the velocity of the combined mass just after collision considering perfectly inelastic collision. (g = 10 m/s²)

Answer 1

15 m/s

Answer 2

To find the velocity of the combined mass just after the perfectly inelastic collision, we need to follow these steps:

1. Determine the time when the two masses collide.
2. Calculate the velocities of both masses just before the collision.
3. Apply the principle of conservation of linear momentum for a perfectly inelastic collision.

Let's set up a coordinate system where the ground is at $y=0$ and the upward direction is positive.

Given:

• Initial height of A, $y_{A0} = 50$ m.
• Initial velocity of A, $u_A = 0$ m/s (released from rest).
• Initial height of B, $y_{B0} = 0$ m.
• Initial velocity of B, $u_B = 50$ m/s (upwards).
• Acceleration due to gravity, $g = 10$ m/s².
• Masses are identical, let $m_A = m_B = m$.

Step 1: Determine the time of collision (t)

The position of mass A at time $t$ is given by:

$y_A(t) = y_{A0} + u_A t - \frac{1}{2}gt^2$

$y_A(t) = 50 + (0)t - \frac{1}{2}(10)t^2$

$y_A(t) = 50 - 5t^2$

The position of mass B at time $t$ is given by:

$y_B(t) = y_{B0} + u_B t - \frac{1}{2}gt^2$

$y_B(t) = 0 + 50t - \frac{1}{2}(10)t^2$

$y_B(t) = 50t - 5t^2$

The collision occurs when $y_A(t) = y_B(t)$:

$50 - 5t^2 = 50t - 5t^2$

$50 = 50t$

$t = 1$ s

So, the collision occurs after 1 second.

Step 2: Calculate the velocities of A and B just before collision

The velocity of an object under constant acceleration is given by $v = u + at$. Here, $a = -g$ (since upward is positive).

For mass A:

$v_A = u_A - gt$

$v_A = 0 - (10)(1)$

$v_A = -10$ m/s (downwards)

For mass B:

$v_B = u_B - gt$

$v_B = 50 - (10)(1)$

$v_B = 40$ m/s (upwards)

Step 3: Apply conservation of linear momentum

For a perfectly inelastic collision, the two masses stick together and move with a common final velocity, $V_f$. The total momentum before the collision equals the total momentum after the collision.

$m_A v_A + m_B v_B = (m_A + m_B) V_f$

Since $m_A = m_B = m$:

$m(-10) + m(40) = (m+m) V_f$

$-10m + 40m = 2m V_f$

$30m = 2m V_f$

$V_f = \frac{30m}{2m}$

$V_f = 15$ m/s

The positive sign for $V_f$ indicates that the combined mass moves upwards just after the collision.

The height at which collision occurs is $y_A(1) = 50 - 5(1)^2 = 45$ m or $y_B(1) = 50(1) - 5(1)^2 = 45$ m.

The velocity of the combined mass just after collision is 15 m/s upwards.

Explanation of the solution:

1. Determine the time of collision by equating the position equations of masses A and B, considering their initial conditions and motion under gravity. This yields $t=1$ s.
2. Calculate the individual velocities of masses A and B just before collision using kinematic equations ($v = u - gt$) at $t=1$ s. $v_A = -10$ m/s (downwards) and $v_B = 40$ m/s (upwards).
3. Apply the principle of conservation of linear momentum for a perfectly inelastic collision: $m_A v_A + m_B v_B = (m_A + m_B) V_f$. Since masses are identical ($m_A=m_B=m$), substitute the values to find the final common velocity $V_f = 15$ m/s.