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Chapter 2: Problem 25

Two small spheres ' \(A\) 'and ' \(B^{\prime}\) mass \(1 \mathrm{~kg}\) and \(2 \mathrm{~kg}\) are moving with speeds \(21 \mathrm{~m} / \mathrm{s}\) and \(4 \mathrm{~m} / \mathrm{s}\) respectively in opposite direction collide head on. After collision \(A\) moves with speed \(1 \mathrm{~m} / \mathrm{s}\) in its initial direction. Which is/are correct? (1) Velocity of \(B\) after collision is \(6 \mathrm{~m} / \mathrm{s}\) opposite to its direction before collision (2) The coefficient of restitution \(e=0.2\) (3) Loss of kinetic energy due to collision is \(200 \mathrm{~J}\) (4) lmpulse of force between two balls is \(400 \mathrm{Ns}\)

Short Answer

Expert verified
The velocity of B after collision is 6 m/s opposite to its initial direction.

Step by step solution

01

Understand the Problem

Two objects, A and B, collide head-on. Their initial speeds are given, and we need to find the velocity of B after the collision, calculate the coefficient of restitution, the loss of kinetic energy, and the impulse.
02

Apply Conservation of Momentum

According to conservation of momentum:\[ m_A u_A + m_B u_B = m_A v_A + m_B v_B \]Substitute the given values:\[ 1 \times 21 + 2 \times (-4) = 1 \times 1 + 2 \times v_B \]This simplifies to:\[ 21 - 8 = 1 + 2v_B \]\[ 13 = 1 + 2v_B \]\[ 12 = 2v_B \]\[ v_B = 6 \text{ m/s} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conservation of Momentum
In physics, conservation of momentum is a fundamental principle that comes into play especially during collisions. It states that the total momentum of a closed system remains constant before and after an event, such as a collision, assuming there is no external force acting on it.

For a head-on collision between two objects like spheres A and B, we can express this concept with the formula:
  • Initial Total Momentum = Final Total Momentum
This means:\[ m_A u_A + m_B u_B = m_A v_A + m_B v_B \] where:
  • \( m_A, m_B \) are the masses of spheres A and B.
  • \( u_A, u_B \) are the initial velocities of spheres A and B.
  • \( v_A, v_B \) are the final velocities of spheres A and B.

In this specific problem, after substituting the given values and solving, sphere B's final velocity comes out to be \( 6 \text{ m/s} \), moving in the opposite direction to its initial movement.
Coefficient of Restitution
The coefficient of restitution \( e \) is a measure of how "bouncy" a collision is. It is the ratio of the relative speeds of separation to the relative speeds of approach. In simple terms, it tells us how much kinetic energy is retained or lost in a collision.
  • A value of \( e = 1 \) implies a perfectly elastic collision where no kinetic energy is lost.
  • A value of \( e = 0 \) indicates a perfectly inelastic collision where objects stick together and kinetic energy is maximally lost.

For this head-on collision:\[ e = \frac{v_B - v_A}{u_A - u_B} \]Substitute the given velocities:
  • After collision: \( v_A = 1 \text{ m/s} \) and \( v_B = 6 \text{ m/s} \) opposite to its original direction.
  • Before collision: \( u_A = 21 \text{ m/s} \) and \( u_B = -4 \text{ m/s} \) (since it is opposite).

Thus: \( e = \frac{6 - 1}{21 + 4} = \frac{5}{25} = 0.2 \). This indicates the collision is somewhat inelastic.
Impulse in Collisions
Impulse represents the change in momentum of an object when a force is applied over a time interval. In collisions, impulse is crucial because it accounts for the transfer of momentum between objects.
The impulse \( J \) can be calculated as:\[ J = \Delta p = m (v - u) \]where:
  • \( \Delta p \) is the change in momentum.
  • \( v \) is the final velocity.
  • \( u \) is the initial velocity.

For sphere B, the impulse is:\[ J = 2 \times (6 - (-4)) = 2 \times (6 + 4) = 2 \times 10 = 20 \text{ Ns} \] Thus, the impulse between the two spheres is \( 20 \text{ Ns} \), highlighting how momentum was transferred during the collision.
Kinetic Energy Loss
During collisions, kinetic energy may be lost especially in inelastic collisions. This lost energy is not necessarily "gone"; it often transforms into other forms of energy, like heat or sound.

To calculate the kinetic energy loss in this collision, compare the total initial kinetic energy with the total final kinetic energy. The formula for kinetic energy (K.E.) is:\[ K. E. = \frac{1}{2} m v^2 \] Initial kinetic energy:
  • Spheres A and B: \( K.E._{initial} = \frac{1}{2} (1 \times 21^2) + \frac{1}{2} (2 \times 4^2) = 220.5 \text{ J} + 16 \text{ J} = 236.5 \text{ J} \)
Final kinetic energy:
  • Spheres A and B: \( K.E._{final} = \frac{1}{2} (1 \times 1^2) + \frac{1}{2} (2 \times 6^2) = 0.5 \text{ J} + 36 \text{ J} = 36.5 \text{ J} \)
Loss of K.E.:
  • \( K.E._{loss} = K.E._{initial} - K.E._{final} = 236.5 - 36.5 = 200 \text{ J} \)
This calculation shows that 200 J of kinetic energy was transformed into other forms of energy due to the collision.

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Most popular questions from this chapter

A ball collides with a fixed inclined plane of inclination \(\theta\) after falling through a distance \(h\). If it moves horizontally just after the impact, the coefficient of restitution is (1) \(\tan \theta\) (2) \(\tan ^{2} \theta\) (3) \(\cot \theta\) (4) \(\cot ^{2} \theta\)

A ball falls vertically onto a floor with momentum \(p\), an then bounces repeatedly. If the coefficient of restitution is \(e\) then the total momentum imparted by the ball on the floor till the ball comes to rest is (1) \(p(1+e)\) (2) \(\frac{p}{1-e}\) (3) \(p\left(1+\frac{1}{e}\right)\) (4) \(p\left(\frac{1+e}{1-e}\right)\)

A particle of mass \(m\) makes an elastic head-on collision with a stationary particle of mass \(2 m .\) If initial kinetic energy is moving particle is \(6 J\), then during the impact, (1) the minimum kinetic energy of the system is \(2 f\) (2) the maximum elastic potential energy of the system is \(4 J\) (3) momentum and total energy are conserved at every instant (4) the ratio of kinetic energy to potential energy of the system first decreases and then increases

A steel ball of mass \(2 \mathrm{~m}\) suffers one-dimensional elastic collision with a row of three steel balls, each of mass \(m\). If mass \(2 m\) has collided with velocity \(v\) and the three balls numbered \(1,2,3\) were initially at rest, then after the collision (1) balls 1,2 and 3 would start moving to the right, each \(_{w_{2}}\) velocity \(v / 3\) (2) balls 2 and 3 would start moving to the right, each \(w_{10}\) velocity \(v / 2\) (3) balls 2 and 3 would start moving to the right, each \(w_{2}\) velocity \(v\) (4) ball 1 and ball of mass \(2 m\) would remain at rest

A glass ball collides with a smooth horizontal surface ( \(x z\) plane) with a velocity \(V=a \hat{i}-b \hat{j} .\) If the coefficient of restitution of collision be \(e\), the velocity of the ball just afie the collision will be (1) \(\sqrt{e^{2} a^{2}+b^{2}}\) at angle \(\tan ^{-1}\left(\frac{a}{e b}\right)\) to the vertical (2) \(\sqrt{a^{2}+e^{2} b^{2}}\) at angle \(\tan ^{-1}\left(\frac{a}{e b}\right)\) to the vertical (3) \(\sqrt{a^{2}+\frac{b^{2}}{e^{2}}}\) at angle \(\tan ^{-1}\left(\frac{e a}{b}\right)\) to the vertical (4) \(\sqrt{\frac{a^{2}}{e^{2}}+b^{2}}\) at angle \(\tan ^{-1}\left(\frac{a}{e b}\right)\) to the vertical
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