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Chapter 3: Problem 413

The force acting on a body whose linear momentum changes by $20 \mathrm{kgms}^{-1}\( in \)10 \mathrm{sec}$ is (A) \(2 \mathrm{~N}\) (B) \(20 \mathrm{~N}\) (C) \(200 \mathrm{~N}\) (D) \(0.2 \mathrm{~N}\)

Short Answer

Expert verified
Using the formula \(F = \frac{\Delta p}{\Delta t}\), we can find the force acting on the body when its linear momentum changes by \(20\,\mathrm{kgms}^{-1}\) in \(10\,\mathrm{sec}\). Plugging in the given values, we get \(F = \frac{20\,\mathrm{kgms}^{-1}}{10\,\mathrm{sec}} = 2 \,\mathrm{N}\). Comparing our result to the given options, the correct answer is option (A) \(2 \mathrm{~N}\).

Step by step solution

01

Write down the formula for force

From Newton's second law of motion, we have the formula for force: \[F = \frac{\Delta p}{\Delta t}\]
02

Plug in the given values

Now, we will substitute the given values into the formula: \(\Delta p = 20\,\mathrm{kgms}^{-1}\) and \(\Delta t = 10\,\mathrm{sec}\) \[F = \frac{20\,\mathrm{kgms}^{-1}}{10\,\mathrm{sec}}\]
03

Calculate the force

We will now perform the division operation to find the force: \[F = 2 \,\mathrm{N}\]
04

Choose the correct option

Comparing our result to the given options, we find that the force acting on the body is : (A) \(2 \mathrm{~N}\) So, the correct answer is option (A).

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Most popular questions from this chapter

With what acceleration (a) should a box descend so that a block of mass \(\mathrm{M}\) placed in it exerts a force \((\mathrm{Mg} / 4)\) on the floor of the box? (A) \((4 \mathrm{~g} / 3)\) (B) \((3 \mathrm{~g} / 4)\) (C) \(\mathrm{g} / 4\) (D) \(3 \mathrm{~g}\)

Three Forces \(F_{1}, F_{2}\), and \(F_{3}\) together keep a body in equilibrium. If \(F_{1}=3 \mathrm{~N}\) along the positive \(\mathrm{X}\) - axis, \(\mathrm{F}_{2}=4 \mathrm{~N}\) along the positive Y-axis then the third force \(F_{3}\) is (A) \(5 \mathrm{~N}\) -making an angle \(\theta=\tan ^{-1}(3 / 4)\) with negative \(\mathrm{y}\) -axis (B) \(5 \mathrm{~N}\) -making an angle \(\theta=\tan ^{-1}(4 / 3)\) with negative \(\mathrm{y}\) -axis (C) \(7 \mathrm{~N}\) -making an angle \(\theta=\tan ^{-1}(3 / 4)\) with negative \(\mathrm{y}\) -axis (D) \(7 \mathrm{~N}\) -making an angle \(\theta=\tan ^{-1}(4 / 3)\) with negative \(\mathrm{y}\) -axis

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 car of mass \(1000 \mathrm{~kg}\) travelling at \(32 \mathrm{~m} / \mathrm{s}\) clashes into a rear of a truck of mass \(8000 \mathrm{~kg}\) moving in the same direction with a velocity of \(4 \mathrm{~m} / \mathrm{s}\). After the collision the car bounces with a velocity of \(8 \mathrm{~ms}^{-1}\). The velocity of truck after the impact is \(\mathrm{m} / \mathrm{s}\) (B) 4 (C) 6 (D) 9 (A) 8

A cricket ball of mass \(150 \mathrm{~g}\). is moving with a velocity of $12 \mathrm{~m} / \mathrm{s}$ and is hit by a bat so that the ball is turned back with a velocity of \(20 \mathrm{~m} / \mathrm{s}\). If duration of contact between the ball and the bat is \(0.01 \mathrm{sec}\). The impulse of the force is (A) \(7.4 \mathrm{NS}\) (B) \(4.8 \mathrm{NS}\) (C) \(1.2 \mathrm{NS}\) (D) \(4.7 \mathrm{NS}\)
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