Question

Find the speed of the second particle as a function of time during the collision.

27. A particle of mass 100 g moving at an initial speed u collides with another particle of same mass initially at rest. If the total kinetic energy becomes 0.2 J after the collision, what could be the minimum and the maximum value of u.

Answer 1

u_{min}=2 m/s, u_{max}=2\sqrt{2} m/s

Answer 2

The problem asks for the minimum and maximum values of the initial speed 'u' of the first particle. The request in the prompt "Find the speed of the second particle as a function of time during the collision" cannot be answered with the given information (Q27) as it requires knowledge of the collision force or duration, which is not provided. We will solve Q27.

Given:

• Mass of first particle, $m_1 = 100 \, \text{g} = 0.1 \, \text{kg}$
• Mass of second particle, $m_2 = 100 \, \text{g} = 0.1 \, \text{kg}$
• Initial speed of first particle, $v_{1i} = u$
• Initial speed of second particle, $v_{2i} = 0$
• Total kinetic energy after collision, $KE_f = 0.2 \, \text{J}$

Let $v_{1f}$ and $v_{2f}$ be the final velocities of the first and second particles, respectively.

1. Conservation of Linear Momentum: The total linear momentum of the system is conserved during the collision. $m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$ Since $m_1 = m_2 = m = 0.1 \, \text{kg}$: $m u + m (0) = m v_{1f} + m v_{2f}$ $u = v_{1f} + v_{2f}$ (Equation 1)

2. Total Kinetic Energy After Collision: The total kinetic energy after the collision is given as $0.2 \, \text{J}$. $KE_f = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2$ $0.2 = \frac{1}{2} (0.1) v_{1f}^2 + \frac{1}{2} (0.1) v_{2f}^2$ $0.2 = 0.05 (v_{1f}^2 + v_{2f}^2)$ $v_{1f}^2 + v_{2f}^2 = \frac{0.2}{0.05}$ $v_{1f}^2 + v_{2f}^2 = 4$ (Equation 2)

3. Solving for u: From Equation 1, substitute $v_{1f} = u - v_{2f}$ into Equation 2: $(u - v_{2f})^2 + v_{2f}^2 = 4$ $u^2 - 2u v_{2f} + v_{2f}^2 + v_{2f}^2 = 4$ $2v_{2f}^2 - 2u v_{2f} + (u^2 - 4) = 0$

This is a quadratic equation in $v_{2f}$. For $v_{2f}$ to have real solutions, the discriminant ($\Delta$) must be non-negative ($\Delta \ge 0$). For a quadratic equation $ax^2 + bx + c = 0$, the discriminant is $\Delta = b^2 - 4ac$. Here, $a=2$, $b=-2u$, $c=(u^2 - 4)$. $\Delta = (-2u)^2 - 4(2)(u^2 - 4) \ge 0$ $4u^2 - 8(u^2 - 4) \ge 0$ $4u^2 - 8u^2 + 32 \ge 0$ $-4u^2 + 32 \ge 0$ $32 \ge 4u^2$ $u^2 \le 8$

Since $u$ is a speed, $u \ge 0$. So, $0 \le u \le \sqrt{8} = 2\sqrt{2}$.

4. Physical Constraint: Energy Loss in Collision: In any collision, the total kinetic energy after the collision ($KE_f$) cannot be greater than the initial total kinetic energy ($KE_i$). $KE_f \le KE_i$ The initial kinetic energy of the system is: $KE_i = \frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 = \frac{1}{2} (0.1) u^2 + \frac{1}{2} (0.1) (0)^2 = 0.05 u^2$

So, $0.2 \le 0.05 u^2$ $u^2 \ge \frac{0.2}{0.05}$ $u^2 \ge 4$ Since $u \ge 0$, we have $u \ge \sqrt{4} = 2$.

5. Combining the Conditions: We have two conditions for $u$:

1. $u^2 \le 8 \implies u \le 2\sqrt{2}$
2. $u^2 \ge 4 \implies u \ge 2$

Combining these, the possible range for $u$ is $2 \le u \le 2\sqrt{2}$.

Minimum and Maximum Values of u:

• The minimum value of $u$ is $u_{min} = 2 \, \text{m/s}$. This occurs when $KE_f = KE_i$, which corresponds to an elastic collision ($e=1$). If $u=2 \, \text{m/s}$, $KE_i = 0.05 (2^2) = 0.2 \, \text{J}$. Since $KE_f = 0.2 \, \text{J}$, this is consistent.
• The maximum value of $u$ is $u_{max} = 2\sqrt{2} \, \text{m/s}$. This occurs when the maximum possible kinetic energy is lost for a given $KE_f$, which corresponds to a perfectly inelastic collision ($e=0$). If $u=2\sqrt{2} \, \text{m/s}$, $KE_i = 0.05 (2\sqrt{2})^2 = 0.05 (8) = 0.4 \, \text{J}$. In this case, $KE_f = 0.2 \, \text{J}$, so half of the initial kinetic energy is lost.

The minimum value of u is $2 \, \text{m/s}$ and the maximum value of u is $2\sqrt{2} \, \text{m/s}$.

The initial prompt's request for "speed of the second particle as a function of time during the collision" cannot be determined from the information provided in Q27, as it requires details about the interaction force during the collision.

The final answer is $\boxed{u_{min}=2 \text{ m/s}, u_{max}=2\sqrt{2} \text{ m/s}}$.

Explanation of the solution:

1. Momentum Conservation: Apply the principle of conservation of linear momentum to the two-particle system. $m u = m v_{1f} + m v_{2f} \implies u = v_{1f} + v_{2f}$
2. Final Kinetic Energy: Use the given total kinetic energy after collision. $\frac{1}{2} m v_{1f}^2 + \frac{1}{2} m v_{2f}^2 = 0.2 \implies v_{1f}^2 + v_{2f}^2 = 4$
3. Quadratic Equation and Discriminant: Substitute $v_{1f}$ from the momentum equation into the kinetic energy equation to get a quadratic equation in terms of $v_{2f}$ and $u$. For real solutions of $v_{2f}$, the discriminant of this quadratic equation must be non-negative. This gives the upper bound for $u$: $u^2 \le 8$.
4. Energy Conservation Principle: The final kinetic energy cannot exceed the initial kinetic energy ($KE_f \le KE_i$). Calculate $KE_i = \frac{1}{2} m u^2 = 0.05 u^2$. This condition gives the lower bound for $u$: $u^2 \ge 4$.
5. Range of u: Combine the two bounds to find the range for $u$: $2 \le u \le 2\sqrt{2}$. The minimum and maximum values are the boundaries of this range.

Answer:

The minimum value of u is $2 \, \text{m/s}$. The maximum value of u is $2\sqrt{2} \, \text{m/s}$.