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Chapter 2: Problem 19

A body of mass \(3 \mathrm{~kg}\) collides elastically with another body at rest and then continues to move in the original direction with one half of its original speed. What is the mass of the target body? (1) \(1 \mathrm{~kg}\) (2) \(1.5 \mathrm{~kg}\) (3) \(2 \mathrm{~kg}\) (4) \(5 \mathrm{~kg}\)

Short Answer

Expert verified
The mass of the target body is 1 kg.

Step by step solution

01

Understand the problem

An elastic collision means both momentum and kinetic energy are conserved. We're given a moving body's initial mass and speed information, and we're asked to find the mass of the second body, which was initially at rest.
02

Conservation of Momentum Equation

The momentum before the collision must equal the momentum after the collision. Let's denote the initial velocity of the 3 kg mass as \(v_0\), and the final velocities of the 3 kg body and the target body as \(v_{f1}\) and \(v_{f2}\) respectively. The momentum equation is:\[ 3\cdot v_0 = 3\cdot v_{f1} + m\cdot v_{f2} \]Where \(m\) is the mass of the second body.
03

Apply Given Condition on Velocities

We're told that after the collision, the 3 kg mass continues with half its original speed, thus \(v_{f1} = \frac{1}{2}v_0\). Substitute this into the momentum equation:\[ 3\cdot v_0 = 3\cdot \frac{1}{2}v_0 + m\cdot v_{f2} \]This simplifies to:\[ \frac{3}{2}v_0 = m\cdot v_{f2} \]
04

Conservation of Kinetic Energy Equation

Since kinetic energy is conserved, set the initial kinetic energy equal to the final kinetic energies:\[ \frac{1}{2} \cdot 3 \cdot v_0^2 = \frac{1}{2} \cdot 3 \cdot (\frac{1}{2}v_0)^2 + \frac{1}{2} \cdot m \cdot v_{f2}^2 \]This simplifies to:\[ \frac{3}{2}v_0^2 = \frac{3}{8}v_0^2 + \frac{1}{2}m v_{f2}^2 \].
05

Solve for Mass of the Target Body

From the equations in steps 3 and 4, solve for \(m\). The momentum equation gives:\[ v_{f2} = \frac{3}{2m}v_0 \].Substitute \(v_{f2}\) into the kinetic energy equation and solve for \(m\):\[ \frac{3}{2}v_0^2 = \frac{3}{8}v_0^2 + \frac{1}{2}m\left(\frac{3}{2m}v_0\right)^2 \], simplify to\[ \frac{3}{2}v_0^2 = \frac{3}{8}v_0^2 + \frac{9}{8}v_0^2 \], leading to \(m = 1.0 \text{ kg}\).

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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, the principle of conservation of momentum is key when analyzing collisions. This principle states that the total momentum of a closed system remains constant if no external forces act upon it. Momentum is simply the product of an object's mass and its velocity. During an elastic collision, like the one presented in our exercise, both momentum and kinetic energy are conserved. This implies that:
  • The momentum before the collision equals the momentum after it.
  • The equation used is: \( m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \).
Here, \(m_1\) and \(m_2\) are the masses of the two colliding bodies, and \(v_{1i}, v_{2i}, v_{1f},\) and \(v_{2f}\) are their initial and final velocities respectively.
In our exercise, we started with a 3 kg body moving with velocity \(v_0\), colliding with a second body initially at rest. After the collision, the 3 kg body retains half of its original velocity, necessitating that the momentum equation be used to solve for the unknown mass of the second body.
Conservation of Kinetic Energy
In elastic collisions, not only is momentum conserved, but so is kinetic energy. Kinetic energy is the energy that a body possesses due to its motion, and is calculated using the equation:
  • \( KE = \frac{1}{2} m v^2 \).
During an elastic collision, the total initial kinetic energy equals the total final kinetic energy, which we can express as:
  • \( \frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2 \).
In our exercise, the initial kinetic energy originates from only one moving mass (3 kg) because the other body is initially at rest. After the collision, we have contributions from both the 3 kg body and the previously stationary body. To find the mass of the target body, it鈥檚 crucial to solve the kinetic energy equation alongside the momentum equation.
Collision Problems
Collision problems in physics can be complex, but breaking them down into fundamental principles can simplify solving them. There are two primary types of collisions鈥攅lastic and inelastic. Our exercise focuses on elastic collisions where:
  • Both momentum and kinetic energy are conserved.
  • You need to use two sets of equations to solve for unknown variables.
The process typically involves:
  • Applying the conservation of momentum to find relations between different velocities and masses.
  • Applying the conservation of kinetic energy to establish another equation.
  • Solving the system of equations to find unknown quantities like mass or velocity.
In our problem, by first understanding the fundamental characteristics of elastic collisions, we systematically apply these principles to calculate the unknown mass of the stationary body based on its interaction with the moving body.

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Most popular questions from this chapter

A ball of mass \(1 \mathrm{~kg}\) is dropped from a height of \(3.2 \mathrm{~m}\) on smooth inclined plane. The coefficient of restitution for the collision is \(e=1 / 2\). The ball's velocity become horizontal after the collision. (1) The angle \(\theta=\tan ^{-1}\left(\frac{1}{\sqrt{2}}\right)\). (2) The speed of the ball after the collision \(=4 \sqrt{2} \mathrm{~ms}^{-1}\). (3) The total loss in kinetic energy during the collision is \(8 \mathrm{~J}\) (4) The ball hits the inclined plane again while travelling vertically downward.

A body \(X\) with a momentum \(p\) collides with another identical stationary body \(Y\) one dimensionally. During the collision, gives an impulse \(J\) to body \(X\). Then coefficient of restitution is (1) \(\frac{2 J}{p}-1\) (2) \(\frac{J}{p}+1\) (3) \(\frac{J}{p}-1\) (4) \(\frac{J}{2 p}-1\)

Two balls \(A\) \& \(B\) of mass \(m_{1}\) and \(m_{2}\) are kept on a horizontal smooth surface. \(A\) is given a velocity towards \(B\) so that they perform head on collision. (1) If \(m_{1}=m_{2}\) and collision is elastic \(A\) stops and \(B\) moves with the velocity of \(A\) after collision (2) Impulses on \(A\) and \(B\) during collision are same in magnitude whether it is elastic or inelastic collision (3) Impulses on \(A\) and \(B\) during collision are equal in magnitude only if \(m_{1}=m_{2}\) (4) If \(m_{1} \gg m_{2}\), though impulses are same in magnitude velocity of \(\bar{A}\) is still in the same direction

A block of mass ' \(m\) ' is hanging from a massless spring of spring constant \(k\). It is in equilibrium under the influence of gravitational force. Another particle of same mass ' \(m\) ' moving upwards with velocity \(u_{0}\) hits the block and sticks to it. For the subsequent motion, choose the incorrect statements: (1) Velocity of the combined mass must be maximum at natural length of the spring. (2) Velocity of the combined mass must be maximum at the new equilibrium position. (3) Velocity of the combined mass must be maximum at the instant particle hits the block. (4) Velocity of the combined mass must be maximum at a point lying between old equilibrium position and natural length.

A bullet of mass \(0.01 \mathrm{~kg}\) and travelling at a speed of \(500 \mathrm{~ms}^{-1}\) strikes a block of mass \(2 \mathrm{~kg}\) which is suspended by a string of length \(5 \mathrm{~m}\). The centre of gravity of the block is found to raise a vertical distance of \(0.2 \mathrm{~m}\). What is the speed of the bullet after it emerges from the block? (1) \(15 \mathrm{~ms}^{-1}\) (2) \(20 \mathrm{~ms}^{-1}\) (3) \(100 \mathrm{~ms}^{-1}\) (4) \(50 \mathrm{~ms}^{-1}\)
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