Question

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

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

Answer 1

None of the options are correct. The calculated value is $\sqrt{\frac{3}{2}}$

Answer 2

To solve this problem, we need to follow these 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: (downward is negative)
  $v_1 = u_1 + at_c = 0 - gt_c = -g\sqrt{\frac{h}{2g}} = -\sqrt{\frac{g^2h}{2g}} = -\sqrt{\frac{gh}{2}}$

• Velocity of Particle 2 ($v_2$) just before collision: (upward is positive)
  $v_2 = u_2 + at_c = \sqrt{2gh} - gt_c = \sqrt{2gh} - g\sqrt{\frac{h}{2g}}$
  $v_2 = \sqrt{2gh} - \sqrt{\frac{g^2h}{2g}} = \sqrt{2gh} - \sqrt{\frac{gh}{2}}$
  We can write $\sqrt{2gh} = \sqrt{4 \cdot \frac{gh}{2}} = 2\sqrt{\frac{gh}{2}}$.
  So, $v_2 = 2\sqrt{\frac{gh}{2}} - \sqrt{\frac{gh}{2}} = \sqrt{\frac{gh}{2}}$

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 $y = y_0 + v_0t + \frac{1}{2}at^2$: (taking ground as $y=0$ and upward as positive, $a=-g$)

$0 = y_c + V t_f + \frac{1}{2}(-g)t_f^2$
$0 = \frac{3h}{4} + 0 \cdot t_f - \frac{1}{2}gt_f^2$
$\frac{1}{2}gt_f^2 = \frac{3h}{4}$
$gt_f^2 = \frac{3h}{2}$
$t_f^2 = \frac{3h}{2g}$
$t_f = \sqrt{\frac{3h}{2g}}$

The question asks for the time in units of $\sqrt{\frac{h}{g}}$.

$t_f = \sqrt{\frac{3}{2}} \sqrt{\frac{h}{g}}$

The value is $\sqrt{\frac{3}{2}}$.

Upon comparing this result with the given options:

(1) $\frac{1}{2}$ (2) $\frac{1}{\sqrt{2}}$ (3) $\sqrt{\frac{4}{3}}$ (4) $\sqrt{\frac{2}{3}}$

None of the options exactly match $\sqrt{\frac{3}{2}}$. This indicates a potential error in the question's options. However, if forced to choose the closest, $\sqrt{\frac{3}{2}} \approx 1.224$ and $\sqrt{\frac{4}{3}} \approx 1.154$. But this is not a valid method for exact science problems. Based on the rigorous derivation, the value is $\sqrt{\frac{3}{2}}$.