A ball of mass 100 g is projected vertically upwards from the ground with a velocity of 49 m/s. At t

Q: Wait, why doesn't the collision change the acceleration of the Centre of Mass?

According to Newton's Second Law for a system of particles, only *external* forces can change the acceleration of the Centre of Mass ($F_{ext} = M_{total} a_{cm}$). The forces the balls exert on each other during the crash are *internal* action-reaction pairs that cancel each other out perfectly. Since gravity is the only external force, $a_{cm}$ remains $-g$ forever.

Q: What if the masses of the two balls were different?

The beauty of this method is that it still works perfectly! You would just calculate the weighted average for the initial position and velocity using the different masses. For example, if the top ball was $2m$ and the bottom was $m$, the initial COM would be closer to the top ball. The acceleration of the COM would still be $-g$.

Q: Why is the displacement of the COM $-49\text{ m}$ and not $0$?

Displacement is defined as Final Position minus Initial Position. The COM starts at a height of $49\text{ m}$ (exactly halfway between $0$ and $98\text{ m}$). It ends up on the ground at a height of $0\text{ m}$. Therefore, the displacement is $0 - 49 = -49\text{ m}$.

Q: Could we have solved this without using the Centre of Mass concept?

Yes, but it is much longer. You would first find the collision time by setting $y_1 = y_2$ (which happens at $t=2\text{ s}$). Then, you'd find their velocities at $t=2\text{ s}$, use momentum conservation to find the velocity of the stuck-together mass, and finally use $s = ut + \frac{1}{2}at^2$ again to find the remaining time to fall to the ground. Both methods give $6.53\text{ s}$!

Q: Why do we reject the negative root of the quadratic equation?

The quadratic equation gives us two mathematical times when the parabola of the COM crosses $y=0$. One is in the future ($t = 6.53\text{ s}$), and one is in the past ($t = -1.53\text{ s}$). Since our physical experiment started at $t=0$, the negative time has no physical meaning in our reality.

Question

A ball of mass $100	ext{ g}$ is projected vertically upwards from the ground with a velocity of $49	ext{ m/s}$. At the same time, another identical ball is dropped from a height of $98	ext{ m}$ to fall freely along the same path as that followed by the first ball. After some time, the two balls collide and stick together and finally fall to the ground. Find the time of flight of the masses.

Sigma Percentile · Accepted Answer

### The Trap of the Two-Part Problem

Imagine the scenario: two balls hurtling towards each other. One is fighting gravity, launched upwards at $49	ext{ m/s}$. The other is surrendering to it, dropped from a height of $98	ext{ m}$.

Inevitably, they collide. And not just a gentle bump—they crash and stick together, becoming a single, heavier mass that eventually plummets back to the earth.

If you try to solve this using standard kinematics, you are walking straight into a trap. You would first have to calculate the exact time they collide. Then, you'd need to find their individual velocities right before impact. After that, you'd apply the conservation of momentum to find the new velocity of the combined mass. Finally, you'd set up a whole new kinematic equation for the second phase of the journey.

It is exhausting, error-prone, and frankly, not the elegant way a physicist thinks.

### The Elegance of the Centre of Mass

Enter the Centre of Mass (COM). This concept is like a cheat code for the universe.

When the two balls collide, the forces they exert on each other are massive. But here is the secret: those forces are strictly **internal** to the two-ball system. According to Newton's laws, internal forces cannot change the motion of the system's Centre of Mass.

From the moment the balls are released, right through the violent collision, and all the way until they hit the ground, the Centre of Mass moves as if nothing happened. It is completely oblivious to the collision!

So, instead of tracking two separate balls, we just track this one imaginary point.

### Setting Up the Master Equation

Let's find out where this magical point starts. At $t = 0$, the first ball is at the ground ($y_1 = 0$) and the second is at $98	ext{ m}$ ($y_2 = 98	ext{ m}$). Since their masses are identical, the COM is exactly in the middle.

$$y_{cm} = \frac{m(0) + m(98)}{m + m} = 49	ext{ m}$$

Next, how fast is it moving initially? We use the same weighted average for velocity. The first ball is moving up at $49	ext{ m/s}$, and the second is dropped from rest ($0	ext{ m/s}$).

$$u_{cm} = \frac{m(49) + m(0)}{m + m} = 24.5	ext{ m/s}$$

What about acceleration? The only external force acting on both balls is gravity. Therefore, the acceleration of the COM is simply the acceleration due to gravity.

$$a_{cm} = -g = -9.8	ext{ m/s}^2$$

### The Final Execution

Here is the beautiful part. After the collision, the balls stick together. The combined mass is now physically located exactly at the Centre of Mass!

To find the total time of flight, we just need to find the time it takes for the Centre of Mass to hit the ground. Hitting the ground means the final position is $0$, so the displacement is $-49	ext{ m}$.

We deploy the second equation of motion for the Centre of Mass:

$$s_{cm} = u_{cm}t + \frac{1}{2}a_{cm}t^2$$

Substituting our values:

$$-49 = 24.5t - \frac{1}{2}(9.8)t^2$$

$$-49 = 24.5t - 4.9t^2$$

Let's clean this up by dividing the entire equation by $4.9$:

$$-10 = 5t - t^2$$

Rearranging this into a standard quadratic form gives:

$$t^2 - 5t - 10 = 0$$

This doesn't factor nicely, so we bring out the quadratic formula:

$$t = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-10)}}{2(1)}$$

$$t = \frac{5 \pm \sqrt{25 + 40}}{2} = \frac{5 \pm \sqrt{65}}{2}$$

Since time must be positive, we discard the negative root. The square root of $65$ is approximately $8.06$.

$$t \approx \frac{5 + 8.06}{2} = \frac{13.06}{2} = 6.53	ext{ s}$$

And there it is. By shifting our perspective to the Centre of Mass, we bypassed the collision entirely and solved a complex two-part problem with a single, elegant equation.

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

The Kinematic Trap

The Centre of Mass (COM)

Initial Position of COM

Initial Velocity of COM

Acceleration of COM

The Combined Mass Trajectory

The Equation of Motion

Substituting Values

Simplifying the Equation

Solving the Quadratic

The Final Answer

The Sigma Insight: Motion of Centre of Mass

The Trap of the Two-Part Problem

The Elegance of the Centre of Mass

Setting Up the Master Equation

The Final Execution

Similar Questions

Comprehension Passage

The value of t is ___ .

The value of x is ___ .

Two identical particles move towards each other with velocity $2 v$ and $v$ , respectively. The velocity of centre of mass is

Two blocks of masses $10 kg$ and $4 kg$ are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of $14 m/s$ to the heavier block in the direction of the lighter block. The velocity of the centre of mass is

Two particles $A$ and $B$ initially at rest, move towards each other by mutual force of attraction. At the instant when the speed of $A$ is $v$ and the speed of $B$ is $2 v$ , the speed of the centre of mass of the system is

A body $A$ of mass $M$ while falling vertically downwards under gravity breaks into two parts; a body $B$ of mass $\frac{1}{3} M$ and a body $C$ of mass $\frac{2}{3} M$ . The centre of mass of bodies $B$ and $C$ taken together shifts compared to that of body $A$ towards

Two masses $A$ and $B$ , each of mass $M$ are fixed together by a massless spring. A force acts on the mass $B$ as shown in figure. If the mass $A$ starts moving away from mass $B$ with acceleration $a$ , then the acceleration of mass $B$ wil be