A point mass of 1kg collides elastically with a stationary point mass of 5kg. After their collision, the 1kg mass reverses its direction and moves with a speed of 2 ms–1. Which of the following statement(s) is (are) correct for the system of these two masses?

Engineering

Physics

Center of Mass

Collision

Momentum and Energy Conservation for System of Particles

Question

A point mass of 1kg collides elastically with a stationary point mass of 5kg. After their collision, the 1kg mass reverses its direction and moves with a speed of 2 ms^–1. Which of the following statement(s) is (are) correct for the system of these two masses?

Kinetic energy of the centre of mass is 0.75 J

Momentum of 5kg mass after collision is 4kg ms^–1

Total kinetic energy of the system is 4J

Total momentum of the system is 3kg ms^–1

v = 5v₂ – 2 (1)

v = v₂ + 92

--------------

2v = 6 v₂

v₂ = v/3

v = 5v/3 – 2

–2v / 3 = – 2

v = 3 m/s

v₂ = 1 m/s

$∴$ P_i = 3 kgms^–1

^{$k_{cm} = \frac{1}{2} (1 + 5) v_{cm}^{2} = 3 v_{cm}^{2}$}

^{$v_{cm} = \frac{1 \times 3}{1 + 5} = \frac{1}{2} \Rightarrow km = \frac{3}{4} = 0 .75$}

^{$\frac{{GM}_{p}}{R_{p}^{2}} = \frac{\sqrt{6}}{11} \times \frac{{GM}_{e}}{R_{e}^{2}} \Rightarrow M_{p} = \frac{\sqrt{6}}{11} \frac{R_{p}^{2}}{R_{e}^{2}} M_{e}$}

^{$\frac{M_{p}}{R_{p}^{3}} = \frac{2}{3} \frac{M_{e}}{R_{e}^{2}} = 1 M_{p} = \frac{2}{3} \frac{M_{e}}{R_{e}^{2}} \times R_{p}^{3}$}

^{$\frac{2}{3} \frac{M_{e}}{R_{e}^{3}} \times R_{p}^{3} = \frac{\sqrt{6}}{11} \times \frac{R_{p}^{2}}{R_{e}^{2}} \times M_{e}$}

^{$R_{p} = \frac{3 \sqrt{6}}{22} R_{e}$}

^{$\frac{v}{11} = \sqrt{\frac{\sqrt{6}}{11} \times \frac{3 \sqrt{6}}{22}} \Rightarrow v = 11 \times \frac{. 1}{11} \times \sqrt{\frac{6 \times 3}{2}} = 3$}

This problem involves an elastic collision between two masses where one is initially moving and the other is stationary. After collision, the smaller mass reverses direction. We need to verify which statements about kinetic energy and momentum are correct.

Step 1: Define Known Quantities

Let mass m₁ = 1 kg, initial velocity u₁ = u (unknown, to be found), final velocity v₁ = -2 m/s (negative since direction reversed).

Mass m₂ = 5 kg, initial velocity u₂ = 0, final velocity v₂ = ? (unknown).

Step 2: Apply Conservation of Momentum

Total initial momentum = Total final momentum

m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}

Substitute values:

(1) u + (5) (0) = (1) (- 2) + (5) v_{2}

Simplifies to:

u = - 2 + 5 v_{2}

Equation (1): $u = 5 v_{2} - 2$

Step 3: Apply Conservation of Kinetic Energy (Elastic Collision)

Total initial K.E. = Total final K.E.

\frac{1}{2} m_{1} {u_{1}}^{2} + \frac{1}{2} m_{2} {u_{2}}^{2} = \frac{1}{2} m_{1} {v_{1}}^{2} + \frac{1}{2} m_{2} {v_{2}}^{2}

Substitute values:

\frac{1}{2} (1) u^{2} + \frac{1}{2} (5) (0) = \frac{1}{2} (1) {(- 2)}^{2} + \frac{1}{2} (5) {v_{2}}^{2}

Multiply both sides by 2 to simplify:

u^{2} = 4 + 5 {v_{2}}^{2}

Equation (2): $u^{2} = 5 {v_{2}}^{2} + 4$

Step 4: Solve the Two Equations

From Eq (1): u = 5v₂ - 2. Substitute this into Eq (2):

{(5 v_{2} - 2)}^{2} = 5 {v_{2}}^{2} + 4

Expand the left side:

25 {v_{2}}^{2} - 20 v_{2} + 4 = 5 {v_{2}}^{2} + 4

Subtract (5v₂² + 4) from both sides:

20 {v_{2}}^{2} - 20 v_{2} = 0

Factor:

20 v_{2} (v_{2} - 1) = 0

This gives v₂ = 0 or v₂ = 1 m/s. v₂ = 0 corresponds to no collision, so we take v₂ = 1 m/s.

Substitute v₂ = 1 into Eq (1) to find u:

u = 5 (1) - 2 = 3

So, initial velocity of 1kg mass, u = 3 m/s.

Step 5: Verify the Given Statements

Statement 1: "Kinetic energy of the centre of mass is 0.75 J"

Velocity of center of mass, v_cm = (m₁u₁ + m₂u₂) / (m₁ + m₂) = (1*3 + 5*0) / (1+5) = 3/6 = 0.5 m/s.

K.E._cm = ½ (m₁ + m₂) v_cm² = ½ * 6 * (0.5)² = 3 * 0.25 = 0.75 J. This statement is CORRECT.

Statement 2: "Momentum of 5kg mass after collision is 4kg ms^–1"

Momentum = m₂v₂ = 5 kg * 1 m/s = 5 kg m/s. The statement says 4 kg m/s. This is INCORRECT.

Statement 3: "Total kinetic energy of the system is 4J"

Total initial K.E. = ½ * 1 * (3)² + ½ * 5 * (0)² = ½ * 9 = 4.5 J.

Total final K.E. = ½ * 1 * (-2)² + ½ * 5 * (1)² = ½ * 4 + ½ * 5 = 2 + 2.5 = 4.5 J.

The total kinetic energy is 4.5 J, not 4 J. This statement is INCORRECT.

Statement 4: "Total momentum of the system is 3kg ms^–1"

Total initial momentum = m₁u₁ + m₂u₂ = 1*3 + 5*0 = 3 kg m/s.

Total final momentum = m₁v₁ + m₂v₂ = 1*(-2) + 5*(1) = -2 + 5 = 3 kg m/s.

Momentum is conserved and equals 3 kg m/s. This statement is CORRECT.

Final Answer: Statements 1 and 4 are correct.

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