Question

Two particles, of masses $M$ and $2M$, moving, as shown, with speeds of $10\, m/s$ and $5\, m/s$, collide elastically at the origin. After the collision, they move along the indicated directions with speeds $\upsilon_1$ and $\upsilon_2$, respectively. The values of $\upsilon_1$ and $\upsilon_2$ are nearly :

• A. 3.2 m/s and 6.3 m/s
• B. 3.2 m/s and 12.6 m/s
• C. 6.5 m/s and 6.3 m/s
• D. 6.5 m/s and 3.2 m/s

Answer 1

The Correct Option is C

To solve this problem, we need to apply the principles of elastic collision. Given that two particles collide elastically, we must ensure both conservation of momentum and conservation of kinetic energy.

Let's denote the masses of the particles as M and 2M. Their initial velocities are 10 \, \text{m/s} and 5 \, \text{m/s}, respectively. After the collision, their velocities are \upsilon_1 and \upsilon_2.

Step 1: Apply Conservation of Momentum

The total momentum before and after the collision must be equal.

The initial momentum \( p_{\text{initial}} \) is given by:

p_{\text{initial}} = M \cdot 10 + 2M \cdot 5 = 10M + 10M = 20M

The final momentum \( p_{\text{final}} \) is given by:

p_{\text{final}} = M \cdot \upsilon_1 + 2M \cdot \upsilon_2

Setting initial momentum equal to final momentum:

M \cdot \upsilon_1 + 2M \cdot \upsilon_2 = 20M

Dividing the entire equation by M, we get:

\upsilon_1 + 2\upsilon_2 = 20 \quad \text{(Equation 1)}

Step 2: Apply Conservation of Kinetic Energy

The total kinetic energy before and after the collision must also be equal.

The initial kinetic energy \( KE_{\text{initial}} \) is:

KE_{\text{initial}} = \frac{1}{2} M (10)^2 + \frac{1}{2} (2M) (5)^2 = 50M + 25M = 75M

The final kinetic energy \( KE_{\text{final}} \) is:

KE_{\text{final}} = \frac{1}{2} M \cdot \upsilon_1^2 + \frac{1}{2} (2M) \cdot \upsilon_2^2

= \frac{M}{2} \upsilon_1^2 + M \upsilon_2^2

Setting initial kinetic energy equal to final kinetic energy:

\frac{M}{2} \upsilon_1^2 + M \upsilon_2^2 = 75M

Dividing the entire equation by M, we get:

\frac{1}{2} \upsilon_1^2 + \upsilon_2^2 = 75 \quad \text{(Equation 2)}

Step 3: Solve the System of Equations

From Equation 1: \upsilon_1 = 20 - 2\upsilon_2

Substitute \upsilon_1 in Equation 2:

\frac{1}{2} (20 - 2\upsilon_2)^2 + \upsilon_2^2 = 75

Simplify and solve the quadratic equation:

200 - 40\upsilon_2 + 2\upsilon_2^2 + \upsilon_2^2 = 75

3\upsilon_2^2 - 40\upsilon_2 + 125 = 0

Using the quadratic formula, we find the solution for \upsilon_2:

\upsilon_2 = \frac{-(-40) \pm \sqrt{(-40)^2 - 4 \cdot 3 \cdot 125}}{2 \cdot 3}

Calculate to get:

\upsilon_2 \approx 6.3 \, \text{m/s}

Substitute \upsilon_2 \approx 6.3 into Equation 1 to get \upsilon_1:

\upsilon_1 = 20 - 2 \times 6.3 = 20 - 12.6 = 7.4 \, \text{m/s}