Question

A particle of mass m is dropped from a height h above the ground. At the same time another of the same mass is thrown vertically upwards from the ground with a speed of $\sqrt{2gh}$. If they head-on completely inelastically, the time taken for the combined mass to reach the ground: $\sqrt{\frac{h}{g}}$ of is:

• A. $\frac{1}{2}$
• B. $\sqrt{\frac{1}{2}}$
• C. $\frac{3}{4}\sqrt{2}$
• D. $\frac{3}{2}\sqrt{2}$

Answer 1

The correct answer is $\sqrt{\frac{3}{2}}$, which is not among the provided options. There may be an error in the question or the options.

Answer 2

To solve this problem, we need to break it down into several steps:

1. Determine the time and position of the collision.
2. Calculate the velocities of both particles just before the collision.
3. Apply the principle of conservation of momentum to find the velocity of the combined mass immediately after the inelastic collision.
4. Calculate the time taken for the combined mass to reach the ground from the collision point.

Let's set up a coordinate system with the ground as the origin ($y=0$) and the upward direction as positive. The acceleration due to gravity is $a = -g$.

Step 1: Time and position of collision

• Particle 1 (dropped from height h):
  Initial position $y_{1,0} = h$
  Initial velocity $u_1 = 0$
  Position at time $t$: $y_1(t) = h - \frac{1}{2}gt^2$

• Particle 2 (thrown vertically upwards from ground):
  Initial position $y_{2,0} = 0$
  Initial velocity $u_2 = \sqrt{2gh}$
  Position at time $t$: $y_2(t) = (\sqrt{2gh})t - \frac{1}{2}gt^2$

Let $t_c$ be the time of collision and $y_c$ be the height of collision. At collision, $y_1(t_c) = y_2(t_c)$:
$h - \frac{1}{2}gt_c^2 = (\sqrt{2gh})t_c - \frac{1}{2}gt_c^2$
$h = (\sqrt{2gh})t_c$
$t_c = \frac{h}{\sqrt{2gh}} = \frac{h}{\sqrt{2}\sqrt{g}\sqrt{h}} = \frac{\sqrt{h}}{\sqrt{2g}} = \sqrt{\frac{h}{2g}}$

Now, calculate the height of collision $y_c$:
$y_c = h - \frac{1}{2}gt_c^2 = h - \frac{1}{2}g\left(\sqrt{\frac{h}{2g}}\right)^2 = h - \frac{1}{2}g\left(\frac{h}{2g}\right) = h - \frac{h}{4} = \frac{3h}{4}$
So, the collision occurs at a height of $\frac{3h}{4}$ from the ground.

Step 2: Velocities just before collision

• Velocity of Particle 1 ($v_1$) just before collision:
  $v_1 = u_1 + at_c = 0 - gt_c = -g\sqrt{\frac{h}{2g}} = -\sqrt{\frac{g^2h}{2g}} = -\sqrt{\frac{gh}{2}}$ (negative sign indicates downward direction)

• Velocity of Particle 2 ($v_2$) just before collision:
  $v_2 = u_2 + at_c = \sqrt{2gh} - gt_c = \sqrt{2gh} - g\sqrt{\frac{h}{2g}}$
  $v_2 = \sqrt{2gh} - \sqrt{\frac{g^2h}{2g}} = \sqrt{2gh} - \sqrt{\frac{gh}{2}}$
  $v_2 = \sqrt{gh}\left(\sqrt{2} - \frac{1}{\sqrt{2}}\right) = \sqrt{gh}\left(\frac{2-1}{\sqrt{2}}\right) = \frac{\sqrt{gh}}{\sqrt{2}} = \sqrt{\frac{gh}{2}}$ (positive sign indicates upward direction)

Step 3: Velocity of combined mass after inelastic collision

Since the collision is completely inelastic, the two particles stick together and move as a single combined mass ($m+m=2m$). Let $V$ be the common velocity after collision. By conservation of momentum:
$m_1v_1 + m_2v_2 = (m_1+m_2)V$
$m\left(-\sqrt{\frac{gh}{2}}\right) + m\left(\sqrt{\frac{gh}{2}}\right) = (m+m)V$
$0 = 2mV$
$V = 0$
The combined mass momentarily comes to rest immediately after the collision.

Step 4: Time taken for the combined mass to reach the ground

The combined mass is at height $y_c = \frac{3h}{4}$ and its initial velocity after collision is $V=0$. It will now fall freely under gravity.
Let $t_f$ be the time taken for the combined mass to reach the ground from this point.
Using the equation of motion $y = y_0 + v_0t + \frac{1}{2}at^2$:
$0 = y_c + V t_f + \frac{1}{2}(-g)t_f^2$
$0 = \frac{3h}{4} + 0 \cdot t_f - \frac{1}{2}gt_f^2$
$\frac{1}{2}gt_f^2 = \frac{3h}{4}$
$gt_f^2 = \frac{3h}{2}$
$t_f^2 = \frac{3h}{2g}$
$t_f = \sqrt{\frac{3h}{2g}} = \sqrt{\frac{3}{2}}\sqrt{\frac{h}{g}}$

The question asks for "the time taken for the combined mass to reach the ground: $\sqrt{\frac{h}{g}}$ of is:". This implies we need to find the factor multiplying $\sqrt{\frac{h}{g}}$.
The factor is $\sqrt{\frac{3}{2}}$.