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Chapter 5: Problem 5

A particle of mass \(m\) describes a circle of radius \(r\). The centripetal acceleration of the particle is \(4 / r^{2}\). The momentum of the particle is (A) \(\frac{4 m}{r}\) (B) \(\frac{2 m}{r}\) (C) \(\frac{4 m}{\sqrt{r}}\) (D) \(v\)

Short Answer

Expert verified
The correct answer is (C) \(\frac{4 m}{\sqrt{r}}\).

Step by step solution

01

The centripetal acceleration, \(a_c = \frac{v^2}{r}\). Also, we are given that \(a_c = \frac{4}{r^2}\). #Step 2: Set the given acceleration equal to the acceleration formula and solve for v#

We have \(\frac{v^2}{r} = \frac{4}{r^2}\). To find the velocity, v, we need to rearrange the equation and solve for v: Multiplying both sides by r, we get: \(v^2 = 4r\) Taking the square root of both sides, we obtain: \(v = 2\sqrt{r}\) #Step 3: Calculate momentum using the velocity found#
02

Now that we have the velocity of the particle, we can find its momentum using the formula \(p = m \times v\): \(p = m(2\sqrt{r})\) #Step 4: Compare the momentum obtained with available options and pick the correct answer

Comparing our expression for momentum with the given options: (A) \(\frac{4 m}{r}\): This does not match our calculation. (B) \(\frac{2 m}{r}\): This does not match our calculation. (C) \(\frac{4 m}{\sqrt{r}}\): This matches our calculation! (D) \(v\): This is not even a momentum expression. So, the correct answer is (C) \(\frac{4 m}{\sqrt{r}}\).

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Most popular questions from this chapter

Vertical circular motion is an example of non-uniform motion and for completion of vertical circular motion the maximum velocity at the start of vertical motion from lowest point is (A) \(\sqrt{g R}\) (B) \(\geq \sqrt{5 g R}\) \((\mathrm{C})=\sqrt{5 g R}\) (D) \(\leq \sqrt{5 g R}\)

A particle of mass \(m\) is placed on top of a smooth hemispherical wedge of mass \(4 m\) at rest. The hemispherical wedge is kept on a frictionless horizontal surface. The particle is given a gentle push. The angular velocity of the particle relative to the wedge if wedge has velocity \(v\) when particle has fallen an angle \(\theta\) with respect to the wedge is (A) \(\frac{5 v}{R \cos \theta}\) (B) \(\frac{v}{R \cos \theta}\) (C) \(\frac{4 v}{R \cos \theta}\) (D) \(\frac{5 v}{R \sin \theta}\)

A particle of mass \(m\) moving eastward with a speed \(v\) collides with another particle of same mass moving northward with same speed \(v .\) The two particles coalesce on collision. The new particle of mass \(2 \mathrm{~m}\) will move in the north-east direction with a velocity of (A) \(v \sqrt{2}\) (B) \(\frac{v}{\sqrt{2}}\) (C) \(\frac{v}{2}\) (D) \(v\)

A block \(Q\) of mass \(M\) is placed on a horizontal frictionless surface \(A B\) and a body \(P\) of mass \(m\) is released on its frictionless slope. As \(P\) slides by a length \(L\) on this slope of inclination \(\theta\), the block \(Q\) would slide by a distance (A) \(\frac{m}{M} L \cos \theta\) (B) \(\frac{m}{M+m} L\) (C) \(\frac{M+m}{m L \cos \theta}\) (D) \(\frac{m L \cos \theta}{m+M}\)

This question has Statement-I and Statement-II. Of the four choices given after the statements, choose the one that best describes the two statements. Statement-I: A point particle of mass \(M\) moving with speed \(v\) collides with stationary point particle of mass \(M\). If the maximum energy loss possible is given as \(f\left(\frac{1}{2} m v^{2}\right)\) then \(f=\left(\frac{m}{M+m}\right)\). Statement-II: Maximum energy loss occurs when the particles get stuck together as a result of the collision. (A) Statement-I is true, Statement-II is true not a correct explanation of Statement-I. (B) Statement-I is true, Statement-II is false. (C) Statement-I is false, Statement-II is true. (D) Statement-I is true, Statement-II is true, Statement-II is a correct explanation of Statement-I.
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