91Ó°ÊÓ

Two balls of masses \(m_{1}\) and \(m_{2}\) are moving towards each other with speeds \(u_{1}\) and \(u_{2}\). They collide head-on and their speeds are \(v_{1}\) and \(v_{2}\) after collision. $\left(m_{1}=8 \mathrm{~kg}, m_{2}=2 \mathrm{~kg}, u_{2}=3 \mathrm{~m} / \mathrm{s}\right)$ Column-I (A) The speed \(u_{1}\) (in \(\mathrm{m} / \mathrm{s}\) ) so that both balls move in same direction if co-efficient of restitution is \(e=0.5\) (B) The speed \(u_{1}\) (in \(\mathrm{m} / \mathrm{s}\) ) so that maximum energy is transformed to \(m_{2}\) (assume elastic collision) (C) Co-efficient of restitution if \(m_{2}\) stops after collision and \(u_{1}=0.5 \mathrm{~m} / \mathrm{s}\) (D) If collision is inelastic and \(u_{1}=3 \mathrm{~m} / \mathrm{s}\), the loss of kinetic energy (in J) after collision may be Column-II 1\. \(\frac{1}{14}\) 2\. \(\frac{1}{8}\) 3\. 2 4\. 4

Step by step solution

01

Conservation of Momentum

The conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. Mathematically, we can write the conservation of momentum for the given problem as \[m_1u_1 - m_2u_2 = m_1v_1 + m_2v_2\]

02

Equation of Restitution

Restitution is a measure of the bounciness of a collision. e, the coefficient of restitution of a collision, is given by: \[e = \frac{Relative\ velocity\ after\ collision}{Relative\ velocity\ before\ collision}\] Now, let's find the value of v1 and v2: \[e = \frac{v_2 - v_1}{u_1 - u_2}\] From the given problem, e=0.5. We plug in the values for e and solve for v1 and v2.

03

Solving for u1

Now, we have two equations with two unknowns, v1 and v2. We can solve this system to find the speeds v1 and v2, and ultimately find the unknown speed u1. On solving, we get \(u_1 = 3 \frac{m}{s}\). Thus, the answer is 3. ##Problem B##

04

Elastic Collision

In an elastic collision, the kinetic energy is conserved. The maximum energy transfer to m2 occurs when m1 stops after the collision, i.e., v1=0, and from the conservation of momentum, v2 = 3u1/m2. Due to the coefficient of restitution, the loss of velocity, e = 1.

05

Solving for u1

Using the principle of conservation of momentum and the equation of restitution, we can solve for u1 and get \(u_1 = 1.5 \frac{m}{s}\). Thus, the answer is 1.5. ##Problem C##

06

Condition for m2 Stopped After Collision

Given that m2 stops after the collision, i.e., v2 = 0, and u1 = 0.5 m/s.

07

Solving for the Coefficient of Restitution

By using the principle of conservation of momentum and equation of restitution, we can solve for the coefficient of restitution and get \(e = \frac{1}{14}\). Thus, the answer is 1/14. ##Problem D##

08

Inelastic Collision

In an inelastic collision, the bodies stick together after the collision and move with a common velocity (v). This implies v1 = v2 = v. Given that u1 = 3 m/s.

09

Solving for Loss of Kinetic Energy

We can find the change in kinetic energies using the conservation of momentum to calculate v and then find the difference between the initial and final kinetic energy: \[Loss\ of\ KE = \frac{1}{2} m_1(u_1^2 - v^2) + \frac{1}{2} m_2(u_2^2 - v^2)\] On solving the equation, we get the loss of kinetic energy as 4 Joules. Thus, the answer is 4.

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