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Chapter 4: Problem 477

A bomb of \(12 \mathrm{~kg}\) divides in two parts whose ratio of masses is $1: 4 .\( If kinetic energy of smaller part is \)288 \mathrm{~J}$, then momentum of bigger part in \(\mathrm{kgm} / \mathrm{sec}\) will be (A) 38 (B) 72 (C) 108 (D) Data is incomplete

Short Answer

Expert verified
The momentum of the bigger part is \(72 \mathrm{~kgm}/\mathrm{sec}\).

Step by step solution

01

Find the masses of the two parts

We are given that the bomb with mass 12 kg divides into two parts. We will find the masses of both parts by using the given ratio 1:4. Divide \(12 \mathrm{~kg}\) in the given ratio: For Smaller part, its mass = \(\frac{1}{(1+4)} \times 12 \mathrm{~kg}\) For Bigger part, its mass = \(\frac{4}{(1+4)} \times 12 \mathrm{~kg}\)
02

Calculate the velocity of the smaller part

The kinetic energy of the smaller part is given as 288 J. To find the velocity of the smaller part, we will use the formula for kinetic energy: \(K.E. = \frac{1}{2}mv^2\) where \(K.E.\) is the kinetic energy, \(m\) is the mass, and \(v\) is the velocity. Rearrange the equation to find velocity: \(v = \sqrt{\frac{2K.E.}{m}}\) Plug in the values, we have: \(v = \sqrt{\frac{2(288 J)}{mass ~of~ small ~part}}\)
03

Use conservation of momentum

According to the conservation of momentum, \(momentum ~before ~ explosion = momentum ~after ~ explosion\) Before the explosion, as the bomb is at rest, its momentum is zero. So, after the explosion, the total momentum is also zero. Smaller part momentum + Bigger part momentum = 0 Calculate the smaller part momentum: \(momentum ~of ~smaller ~part = mass ~of ~smaller ~part \times velocity ~of ~smaller ~part\) So we have, \(momentum ~of ~bigger ~part = -momentum ~of ~smaller ~part\)
04

Calculate the momentum of the bigger part

We can now calculate the momentum of the bigger part by plugging in the values from previous steps: \(momentum ~of ~bigger ~part = -momentum ~of ~smaller ~part = -mass ~of ~smaller ~part \times velocity ~of ~smaller ~part\) Plug in the values, and we have the momentum of the bigger part. Finally, compare the result with the given options to select the correct answer.

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Most popular questions from this chapter

A body of mass \(6 \mathrm{~kg}\) is under a force, which causes a displacement in it given by $\mathrm{S}=\left[\left(2 \mathrm{t}^{3}\right) / 3\right](\mathrm{in} \mathrm{m}) .$ Find the work done by the force in first one seconds. (A) \(2 \mathrm{~J}\) (B) \(3.8 \mathrm{~J}\) (C) \(5.2 \mathrm{~J}\) (D) \(24 \mathrm{~J}\)

A spring gun of spring constant \(90 \times 10^{2} \mathrm{~N} / \mathrm{M}\) is compressed \(4 \mathrm{~cm}\) by a ball of mass \(16 \mathrm{~g}\). If the trigger is pulled, calculate the velocity of the ball. (A) \(60 \mathrm{~m} / \mathrm{s}\) (B) \(3 \mathrm{~m} / \mathrm{s}\) (C) \(90 \mathrm{~m} / \mathrm{s}\) (D) \(30 \mathrm{~m} / \mathrm{s}\)

A spring of spring constant \(10^{3} \mathrm{~N} / \mathrm{m}\) is stretched initially \(4 \mathrm{~cm}\) from the unscratched position. How much the work required to stretched it further by another \(5 \mathrm{~cm}\) ? (A) \(6.5 \mathrm{NM}\) (B) \(2.5 \mathrm{NM}\) (C) \(3.25 \mathrm{NM}\) (D) \(6.75 \mathrm{NM}\)

A spherical ball of mass \(15 \mathrm{~kg}\) stationary at the top of a hill of height \(82 \mathrm{~m} .\) It slides down a smooth surface to the ground, then climbs up another hill of height \(32 \mathrm{~m}\) and finally slides down to horizontal base at a height of \(10 \mathrm{~m}\) above the ground. The velocity attained by the ball is (A) \(30 \sqrt{10 \mathrm{~m} / \mathrm{s}}\) (B) \(10 \sqrt{30 \mathrm{~m} / \mathrm{s}}\) (C) \(12 \sqrt{10} \mathrm{~m} / \mathrm{s}\) (D) \(10 \sqrt{12} \mathrm{~m} / \mathrm{s}\)

A uniform chain of length \(\mathrm{L}\) and mass \(\mathrm{M}\) is lying on a smooth table and \((1 / 4)^{\text {th }}\) of its length is hanging vertically down over the edge of the table. If \(g\) is acceleration due to gravity, the work required to pull the hanging part on to the table is (A) MgL (B) \(\mathrm{MgL} / 9\) (C) \(\mathrm{MgL} / 18\) (D) \(\mathrm{MgL} / 32\)
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