PhysicsWork, energy and power · centre of massClass 11

Collisions, centre of mass and rockets: interactive simulation

In every collision the total momentum stays the same, because the forces the bodies exert on each other are equal and opposite. What changes from one collision to another is how much kinetic energy survives, measured by the coefficient of restitution e.

Useful for: JEE Main, JEE Advanced, NEET, CBSE Boards, AP Physics 1, AP Physics C: Mechanics, A-level Physics, IB Physics · 3 modes in one lab

What you can explore

  1. 1

    1D collision

    Head-on collisions with any masses, velocities and e, with momentum and energy bars and a CM-frame view.
  2. 2

    Oblique (2D) collision

    Off-centre impact, line of centres and the 90° rule for equal elastic masses.
  3. 3

    Rocket (variable mass)

    Thrust, rising acceleration, burn-out speed and the rocket equation.

Try this

Smash two carts head-on and watch momentum and energy bars, view the same collision from the centre of mass, hit a ball off-centre to see an oblique collision, and fire a rocket whose mass keeps falling.

  1. 1

    Collide 2 kg at 4 m/s with 1 kg at −2 m/s, e = 1.

    What you should see

    The 2 kg stops dead and the 1 kg bounces back at 6 m/s. Momentum stays 6 kg m/s and KE stays 18 J.

  2. 2

    Set e = 0 with the same masses and speeds.

    What you should see

    They stick together at v_cm = 2 m/s and 12 J of kinetic energy is lost.

  3. 3

    Turn on the centre-of-mass frame.

    What you should see

    The CM marker stands still; the bodies approach and separate with equal and opposite momenta.

  4. 4

    In Oblique collision, use equal masses with e = 1 and any impact parameter.

    What you should see

    The two balls always leave at 90° to each other.

  5. 5

    In Rocket, use 1000 kg, 25 % dry mass, 10 kg/s and v_r = 2000 m/s.

    What you should see

    Thrust 20 kN, initial acceleration 10.2 m/s², burn time 75 s and burn-out speed about 2040 m/s (2770 m/s without gravity).

Formulas behind it

Conservation of momentum

m1u1+m2u2=m1v1+m2v2m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2

Coefficient of restitution

e=v2−v1u1−u2e = \frac{v_2 - v_1}{u_1 - u_2}

Final velocity of body 1

v1=m1u1+m2u2−m2e(u1−u2)m1+m2v_1 = \frac{m_1u_1 + m_2u_2 - m_2e(u_1 - u_2)}{m_1 + m_2}

Velocity of the centre of mass

vcm=m1u1+m2u2m1+m2v_{cm} = \frac{m_1u_1 + m_2u_2}{m_1 + m_2}

Rocket equation (no gravity)

v=vrln⁡m0mv = v_r\ln\frac{m_0}{m}

Rocket acceleration

a=vrmdmdt−ga = \frac{v_r}{m}\frac{dm}{dt} - g

Common mistakes

  • Conserving kinetic energy in an inelastic collision. Only momentum is conserved in general.
  • Taking speeds instead of signed velocities in the momentum equation.
  • Applying e along the original direction of motion in an oblique collision. It applies along the line of centres.
  • Using F = ma with constant m for a rocket. The mass changes, so the thrust term v_r dm/dt must be included.
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FAQ

When do two bodies simply exchange velocities?

In a perfectly elastic head-on collision between equal masses. The moving one stops and the stationary one moves off with its speed.

Can all the kinetic energy be lost in a collision?

Only when the total momentum is zero (seen from the CM frame). Otherwise the combined body must still move at v_cm.

Does a rocket need air to push against?

No. It pushes the exhaust gases backwards and they push the rocket forwards — Newton’s third law works in a vacuum.

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