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Chapter 8: Problem 13

A body of mass \(M\) moving with velocity \(v \mathrm{~ms}^{-1}\) suddenly breaks into two pieces. One part having mass M/4 remains stationary. The velocity of the other part will be (a) \(\mathrm{v}\) (b) \(2 \mathrm{v}\) (c) \(\frac{3 v}{4}\) (d) \(\frac{4 \mathrm{v}}{3}\)

Short Answer

Expert verified
(d) \( \frac{4v}{3} \)

Step by step solution

01

Understand the Problem

A body of mass \( M \) moving with velocity \( v \text{ m/s} \) breaks into two parts. One portion with mass \( \frac{M}{4} \) stays stationary, while the velocity of the other part must be calculated.
02

Apply Conservation of Momentum

According to the conservation of momentum, the total momentum before and after the separation of the body must remain constant. Initially, the momentum is \( Mv \). After separation, momentum is divided as follows: the part with mass \( \frac{M}{4} \) has no velocity, and the remaining \( \frac{3M}{4} \) portion moves with some velocity \( v_f \).
03

Set Up the Equation

The conservation of momentum gives us the equation:\[ Mv = \frac{M}{4} \times 0 + \frac{3M}{4} \times v_f \].Simplify the right side: \(\frac{3M}{4} \times v_f \).
04

Solve for Final Velocity \(v_f\)

To find \(v_f\), we solve the equation: \[ Mv = \frac{3M}{4} \times v_f \].Cancel out \( M \) from both sides:\[ v = \frac{3}{4}v_f \].Rearrange to solve for \( v_f \):\[ v_f = \frac{4}{3}v \].
05

Verify Solution

The calculations have provided a solution for the velocity of the second part, which is \( \frac{4v}{3} \), matching option (d).

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Key Concepts

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

Inelastic Collision
Inelastic collision events occur when two objects collide and stick together, or when they split apart and do not retain their original momentum individually. In our original exercise, a body breaks into two pieces, yet it鈥檚 considered an inelastic collision because the parts share the total momentum initially held by the whole. When addressing inelastic collisions, it's critical to note:
  • The kinetic energy is not conserved. Instead, it may be transformed into other energy forms, such as heat or deformation work.
  • The total momentum, however, remains conserved. Both pieces from the fracture collectively maintain the initial momentum.
Understanding inelastic collisions helps us grasp how momentum behaves in real-world scenarios, like a car crash, where vehicles may crumple but conserve momentum.
Momentum Calculation
Momentum is a fundamental concept in physics, representing the quantity of motion an object possesses. It's calculated by multiplying mass with velocity: \ (\text{momentum} = \text{mass} \times \text{velocity}) \.
In the exercise provided, the body's initial momentum was calculated as \(Mv\), given by the mass \(M\) and the velocity \(v\) of the entire body.
When the body divides, each section contributes to the overall momentum, calculated individually.
  • The stationary piece has zero velocity, so its momentum contribution is also zero.
  • The moving piece contributes a momentum \(\frac{3M}{4} \times v_f\).
By applying the principle of momentum conservation, these values help determine the final velocity \(v_f\) of the moving part. Always remember, the total initial momentum equals the total final momentum in every collision.
Dynamics of Motion
The dynamics of motion involve the laws governing how objects move, encompassing forces and the conservation principles. In scenarios like this exercise, motion dynamics emphasize how bodies react when exerting internal forces, like breaking.
  • Newton鈥檚 first law highlights that a body will continue its motion unless an external force acts on it.
  • Newton's second law, \(F = ma\), tells us how the velocity of an object changes when a force is applied.
  • The conservation of momentum insists that within a closed system, like our body splitting, the total momentum remains constant.
Investigating dynamics of motion enables us to predict motion results from forces, ensuring we can solve practical problems, such as predicting positions or velocities post-fracture of a moving object.

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Most popular questions from this chapter

Assertion-Retson type. Each of these contains two Statements: Statement I (Assertion), Statement \(\Pi\) (Rotson). Fach of these questions also has fomr alternatitue choice, only ane of uhich is correct. You hatee to select the correct choices from the coves (al, (b), (c) and (d) ginvn bolow (a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion (b) If both Assertion and Reason are true but Reason is not correct explanation of the Assertion (c) If Assertion is true but Reason is false (d) If Assertion is false but the Reason is true Assertion The centre of mass of a two particle system lies on the line joining the two particles, being closer to the heavier particle. Reason This is because produet of mass of one particle and its distance from centre of mass is numerically equal to product of mass of other particle and its distance from centre of mass.

Two boys of masses \(10 \mathrm{~kg}\) and \(8 \mathrm{~kg}\) are moving along a vertical rope. The former climbing up with acceleration of \(2 \mathrm{~ms}^{-1}\) while later coming down with uniform velocity of \(2 \mathrm{~ms}^{-1}\). Then tension in rope at fixed support will be (Take \(g=10 \mathrm{~ms}^{-2}\) ) (a) \(200 \mathrm{~N}\) (b) \(120 \mathrm{~N}\) (c) \(180 \mathrm{~N}\) (d) \(160 \mathrm{~N}\)

A particle of mass \(m\) moving with a velocity \(v\) makes an elastic one dimensional collision with a stationary particle of mass establishing a contact with it for extremely small time \(T\). Their force of contact increase from zero to \(f_{0}\) linearly in time \(T / 4\) remains constant for the further time \(\frac{T}{2}\) and decrease linearly from \(f_{0}\) to zero in further time \(\frac{T}{4}\) as shown. The magnitude possessed by \(f_{0}\) is (a) \(\frac{\mathrm{mv}}{\mathrm{T}}\) (b) \(\frac{2 m v}{T}\) (c) \(\frac{4 \mathrm{mv}}{3 T}\) (d) \(\frac{3 m v}{T}\)

A bullet of mass \(50 \mathrm{~g}\) is fired from a gun of mass \(2 \mathrm{~kg}\). If the total kinetic energy produced is \(2050 \mathrm{~J}\), the kinetic energy of the bullet and the gun respectively are (a) \(200 j, 5\rfloor\) (b) \(2000 \mathrm{~J}, 50 \mathrm{~J}\) (c) \(5 \mathrm{~J}, 200 \mathrm{~J}\) (d) \(50 \mathrm{~J}, 2000 \mathrm{~J}\)

The density of a non-uniform rod of length \(1 \mathrm{~m}\) is given by \(\rho(x)=a\left(1+b x^{2}\right)\) where \(a\) and \(b\) are constants and \(0 \leq x \leq 1\). The centre of mass of the rod will be at |NCERT Exemplar| (a) \(\frac{3(2+b)}{4(3+b)}\) (b) \(\frac{4(2+b)}{3(3+b)}\) (c) \(\frac{3(2+b)}{4(2+b)}\) (d) \(\frac{4(3+b)}{3(2+b)}\)
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