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Chapter 3: Problem 414

An impulsive force of \(100 \mathrm{~N}\) acts on a body for \(1 \mathrm{sec}\) What is the change in its linear momentum ? (A) \(10 \mathrm{~N}-\mathrm{S}\) (B) \(100 \mathrm{~N}-\mathrm{S}\) (C) \(1000 \mathrm{~N}-\mathrm{S}\) (D) \(1 \mathrm{~N}-\mathrm{S}\)

Short Answer

Expert verified
The change in linear momentum is \(100 \mathrm{~N}-\mathrm{S}\).

Step by step solution

01

Recall the impulse-momentum theorem formula

The formula for the impulse-momentum theorem is given as: Impulse = Change in Momentum Impulse = Force × Time Change in Momentum = Force × Time
02

Plug in the given values and calculate the change in momentum

From the problem, the force is 100 N and the time is 1 sec. Plug these values into the formula: Change in Momentum = 100 N × 1 s Change in Momentum = 100 N·s
03

Compare our result with the options

Our calculation gave us a change in momentum of 100 N·s. Compare this result to the options given. The correct answer is: (B) \(100 \mathrm{~N}-\mathrm{S}\)

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Most popular questions from this chapter

A body of mass \(5 \mathrm{~kg}\) starts motion form the origin with an initial velocity $\mathrm{v}_{0} \rightarrow=30 \mathrm{i}+40 \mathrm{j} \mathrm{m} / \mathrm{s}$ If a constant force \(\mathrm{F}=-\left(\mathrm{i}^{\wedge}+5 \mathrm{j}\right) \mathrm{N}\) acts on the body, than the time in which the Y-component of the velocity becomes zero is (A) \(5 \mathrm{~s}\) (B) \(20 \mathrm{~s}\) (C) \(40 \mathrm{~s}\) (D) \(80 \mathrm{~s}\)

A bag of sand of mass \(\mathrm{m}\) is suspended by rope. a bullet of mass \((\mathrm{m} / 30)\) is fired at it with a velocity \(\mathrm{V}\) and gets embedded into it. The velocity of the bag finally is (A) \((31 \mathrm{~V} / 30)\) (B) \((30 \mathrm{~V} / 31)\) (C) \((\mathrm{V} / 31)\) (D) \((\mathrm{V} / 30)\)

A ball of mass \(0.2 \mathrm{~kg}\) is thrown vertically upwards by applying a force by hand. if the hand moves \(0.2 \mathrm{~m}\) while applying the force and the ball goes up to \(2 \mathrm{~m}\) height further. find the magnitude of the force. (Consider \(\mathrm{g}=10 \mathrm{~ms}^{-1}\) ) (A) \(16 \mathrm{~N}\) (B) \(20 \mathrm{~N}\) (C) \(22 \mathrm{~N}\) (D) \(4 \mathrm{~N}\)

Two bodies of equal masses revolve in circular orbits of radii \(\mathrm{R}_{1}\) and \(\mathrm{R}_{2}\) with the same period Their centripetal forces are in the ratio. (A) \(\left(\mathrm{R}_{2} / \mathrm{R}_{1}\right)^{2}\) (B) \(\left(\mathrm{r}_{1} / \mathrm{r}_{2}\right)\) (C) \(\left(\mathrm{R}_{1} / \mathrm{R}_{2}\right)^{2}\) (D) \(\left.\sqrt{(}_{1} R_{2}\right)\)

A body of mass \(0.05 \mathrm{~kg}\) is falling with acceleration $9.4 \mathrm{~ms}^{-2}$. The force exerted by air opposite to motion is \(\mathrm{N}\) \(\left(g=9.8 \mathrm{~ms}^{-2}\right)\) (A) \(0.02\) (B) \(0.20\) (C) \(0.030\) (D) Zero
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