BITSAT PYQs for Centre of Mass with Solutions: Practice BITSAT Previous Year Questions

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Shivam Yadav

Updated on - Dec 12, 2025

Centre of Mass is an important topic in the Physics section in BITSAT exam. Practising this topic will increase your score overall and make your conceptual grip on BITSAT exam stronger.

This article gives you a full set of BITSAT PYQs for Centre of Mass with explanations for effective preparation. Practice of BITSAT Physics PYQs including Centre of Mass questions regularly will improve accuracy, speed, and confidence in the BITSAT 2026 exam.

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BITSAT PYQs for Centre of Mass with Solutions

BITSAT PYQs for Centre of Mass with Solutions

  • 1.
    Two particles $A$ and $B$ initially at rest, move towards each other, under mutual force of attraction. At an instance when the speed of $A$ is $v$ and speed of $B$ is $2v$, the speed of center of mass $(CM)$ is

      • Zero
      • V
      • 2.5v
      • 4v

    • 2.
      A man of mass \(100 \,kg\). is standing on a platform of mass \(200 \,kg\). Which is kept on a smooth ice surface. If the man starts moving on the platform with a speed \(30 \,m/sec\) relative to the platform then calculate with what velocity relative to the ice the platform will recoil?

        • 5 m/sec
        • 10 m/sec
        • 15 m/sec
        • 20 m/sec

      • 3.

        A system consists of three particles, each of mass \(m\) and located at \((1,1),(2,2)\) and \((3,3)\). The co-ordinates of the center of mass are :

          • (1, 1)
          • (2,2)
          • (3,3)
          • (6,6)

        • 4.
          A spherically symmetric gravitational system of particles has a mass density $ \rho = \begin{cases} \rho_0 & \text{for} \; r \le R \\ 0 & \text{for} \; r > R \end{cases}$ where $r_0$ is a constant. A test mass can undergo circular motion under the influence of the gravitational field of particles. Its speed $V$ as a function of distance $r (0 < r < \infty) $ from the centre of the system is represented by


          • 5.
            At any instant the velocity of a particle of mass 500g is \( \left( 2t \hat{i} + 3t^2 \hat{j} \right) \, \text{ms}^{-1} \). If the force acting on the particle at \( t = 1 \) s is \( \left( \hat{i} + x \hat{j} \right) \, \text{N} \), then the value of \( x \) will be:

              • 3
              • 4
              • 6
              • 2

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