A body A is projected with a velocity of 40 ms-1 at an angle of 60° with the horizontal. At some other time, another body B is thrown vertically upwards with a velocity of 40√3 ms-1 such that it collides body A at a height of 60 m. If the velocity of centre of mass of the two bodies after collision is 20 ms-1, then the ratio of the masses of the bodies A and B is (Acceleration due to gravity = 10 ms-2)

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q89 – AP-EAMCET 2026

← Back

Home Physics Motion in a PlaneQuestion #297

A body A is projected with a velocity of

40 {ms}^{- 1}

at an angle of

60^{\circ}

with the horizontal. At some other time, another body B is thrown vertically upwards with a velocity of

40 \sqrt{3} {ms}^{- 1}

such that it collides body A at a height of

60 m

. If the velocity of centre of mass of the two bodies after collision is

20 {ms}^{- 1}

, then the ratio of the masses of the bodies A and B is (Acceleration due to gravity =

10 {ms}^{- 2}

)

Clue 1 of 3: Kinematics of Projectile Motion

First, determine the time at which body A reaches a height of $60 m$ and its velocity (magnitude and direction) at that specific moment. Remember that the vertical component of velocity changes due to gravity, while the horizontal component remains constant.

🧠 Full Solution Path

Step 1: Determine the velocity of body A at the collision height.✦ Active

Body A is projected with an initial velocity $u_{A} = 40 {ms}^{- 1}$ at an angle $θ = 60^{\circ}$ with the horizontal. We first find the time it takes for body A to reach a height of $h = 60 m$ .

y = (u_{A} \sin θ) t - \frac{1}{2} g t^{2}

60 = (40 \sin 60^{\circ}) t - \frac{1}{2} (10) t^{2}

60 = (40 \times \frac{\sqrt{3}}{2}) t - 5 t^{2}

60 = 20 \sqrt{3} t - 5 t^{2}

5 t^{2} - 20 \sqrt{3} t + 60 = 0

t^{2} - 4 \sqrt{3} t + 12 = 0

(t - 2 \sqrt{3})^{2} = 0 ⟹ t = 2 \sqrt{3} s

Now, we find the components of velocity of body A at $t = 2 \sqrt{3} s$ :

v_{A x} = u_{A} \cos θ = 40 \cos 60^{\circ} = 40 \times \frac{1}{2} = 20 {ms}^{- 1}

v_{A y} = u_{A} \sin θ - g t = 20 \sqrt{3} - 10 (2 \sqrt{3}) = 0 {ms}^{- 1}

So, the velocity of body A just before collision is ${\vec{v}}_{A} = 20 \hat{i} {ms}^{- 1}$ .

💡 Teacher's Secret Hint

Note that $v_{A y} = 0$ means body A is at its maximum height when the collision occurs. This simplifies the velocity vector.

Step 2: Determine the velocity of body B at the collision height.○ Expand

Body B is thrown vertically upwards with initial velocity $u_{B} = 40 \sqrt{3} {ms}^{- 1}$ . It collides with body A at a height of $h = 60 m$ . We calculate its speed at this height using the kinematic equation $v^{2} = u^{2} - 2 g y$ .

v_{B}^{2} = u_{B}^{2} - 2 g h

v_{B}^{2} = (40 \sqrt{3})^{2} - 2 (10) (60)

v_{B}^{2} = (1600 \times 3) - 1200

v_{B}^{2} = 4800 - 1200 = 3600

v_{B} = \sqrt{3600} = 60 {ms}^{- 1}

To determine the direction, we consider the time of flight for body B to reach $60 m$ (while ascending). The problem states "At some other time," implying body B's launch time is adjusted to meet body A. For body B to be at $60 m$ moving upwards, the velocity is $+ 60 {ms}^{- 1}$ . This is confirmed by calculating the time $t^{'}$ for B to reach $60 m$ on its way up: $60 = 40 \sqrt{3} t^{'} - \frac{1}{2} (10) t^{' 2}$ , which gives $t^{'} = 4 \sqrt{3} - 6 s$ . The velocity at this time is $v_{B} = u_{B} - g t^{'} = 40 \sqrt{3} - 10 (4 \sqrt{3} - 6) = 60 {ms}^{- 1}$ . Thus, the velocity of body B just before collision is ${\vec{v}}_{B} = 60 \hat{j} {ms}^{- 1}$ .

💡 Teacher's Secret Hint

The phrase "At some other time" implies that the launch time of body B is adjusted such that it meets body A at the specified height and time. This ensures consistency for the collision event.

Step 3: Calculate the ratio of masses using the velocity of the center of mass.○ Expand

The velocity of the center of mass of the system $(m_{A} + m_{B})$ just before collision is given by ${\vec{V}}_{C M} = \frac{m_{A} {\vec{v}}_{A} + m_{B} {\vec{v}}_{B}}{m_{A} + m_{B}}$ . The magnitude of this velocity is given as $20 {ms}^{- 1}$ .

{\vec{V}}_{C M} = \frac{m_{A} (20 \hat{i}) + m_{B} (60 \hat{j})}{m_{A} + m_{B}}

V_{C M, x} = \frac{20 m_{A}}{m_{A} + m_{B}}

V_{C M, y} = \frac{60 m_{B}}{m_{A} + m_{B}}

| {\vec{V}}_{C M} |^{2} = V_{C M, x}^{2} + V_{C M, y}^{2} = (20)^{2}

{(\frac{20 m_{A}}{m_{A} + m_{B}})}^{2} + {(\frac{60 m_{B}}{m_{A} + m_{B}})}^{2} = 400

\frac{400 m_{A}^{2}}{(m_{A} + m_{B})^{2}} + \frac{3600 m_{B}^{2}}{(m_{A} + m_{B})^{2}} = 400

\frac{m_{A}^{2} + 9 m_{B}^{2}}{(m_{A} + m_{B})^{2}} = 1

m_{A}^{2} + 9 m_{B}^{2} = (m_{A} + m_{B})^{2}

m_{A}^{2} + 9 m_{B}^{2} = m_{A}^{2} + 2 m_{A} m_{B} + m_{B}^{2}

8 m_{B}^{2} = 2 m_{A} m_{B}

4 m_{B} = m_{A}

\frac{m_{A}}{m_{B}} = \frac{4}{1}

The ratio of the masses of bodies A and B is $4 : 1$ .

💡 Teacher's Secret Hint

Remember that the velocity of the center of mass remains constant during a collision if no external forces act. In this case, gravity is an external force, but the given $20 {ms}^{- 1}$ is the velocity *of the CM of the system* at the time of collision, which means it represents ${\vec{V}}_{C M}$ just before (and during/after the instant of) collision.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Physics Question 89 – AP-EAMCET 2026

Clue 1 of 3: Kinematics of Projectile Motion

🧠 Full Solution Path