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Chapter 4: Problem 508

Two solid rubber balls \(P\) and \(Q\) having masses \(200 \mathrm{~g}\) and $400 \mathrm{~g}$ respectively are moving in opposite directions with velocity of \(\mathrm{P}\) equal to \(0.3 \mathrm{~m} / \mathrm{s}\). After collision the two balls come to rest, then the velocity of \(Q\) is (A) \(0.15 \mathrm{~m} / \mathrm{s}\) (B) \(1.5 \mathrm{~m} / \mathrm{s}\) (C) \(-0.15 \mathrm{~m} / \mathrm{s}\) (D) Zero

Short Answer

Expert verified
The initial velocity of ball Q is \(-0.15\,\text{m/s}\).

Step by step solution

01

Write the equation for the conservation of linear momentum

Using the principle of conservation of linear momentum, we can write the equation as follows: $$m_Pv_P + m_Qv_Q = (m_P + m_Q)v_{final}$$ Here, \(m_P\) and \(m_Q\) represent the masses of balls P and Q; \(v_P\) and \(v_Q\) represent their initial velocities; and \(v_{final}\) represents the final velocity of both balls after the collision. Since the two balls come to rest after the collision, the final velocity of the system, \(v_{final}\), is zero.
02

Substitute the given values into the equation

Substitute the given values for the masses and velocity of ball P into the equation: $$(200\,\text{g})(0.3\,\text{m/s}) + (400\,\text{g})v_Q = 0$$
03

Solve for the initial velocity of ball Q, \(v_Q\)

Simplify the equation and solve for \(v_Q\): \begin{align*} (200\,\text{g})(0.3\,\text{m/s}) &+ (400\,\text{g})v_Q = 0 \\ (60\,\text{g m/s}) &+ (400\,\text{g})v_Q = 0 \\ 400\,\text{g}v_Q &= -60\,\text{g m/s} \\ v_Q &= \dfrac{-60\,\text{g m/s}}{400\,\text{g}} \end{align*}
04

Calculate the value of \(v_Q\)

Solve for \(v_Q\) and simplify: $$v_Q = \dfrac{-60\,\text{g m/s}}{400\,\text{g}} = -0.15\,\text{m/s}$$ This matches the value given in option (C). So, the initial velocity of ball Q is \(-0.15\,\text{m/s}\).

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Most popular questions from this chapter

\(1 \mathrm{~kg}\) apple gives \(25 \mathrm{KJ}\) energy to a monkey. How much height he can climb by using this energy if his efficiency is \(40 \%\). (mass of monkey \(=25 \mathrm{~kg}\) and \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\) ) (A) \(20 \mathrm{~m}\) (B) \(4 \mathrm{~m}\) (C) \(30 \mathrm{~m}\) (D) \(40 \mathrm{~m}\)

A single conservative force \(\mathrm{F}(\mathrm{x})\) acts on a $2.5 \mathrm{~kg}\( particle that moves along the \)\mathrm{x}$ -axis. The potential energy \(\mathrm{U}(\mathrm{x})\) is given by \(\mathrm{U}(\mathrm{x})=\left[10+(\mathrm{x}-4)^{2}\right]\) where \(\mathrm{x}\) is in meter. \(\mathrm{At} \mathrm{x}=6.0 \mathrm{~m}\) the particle has kinetic energy of \(20 \mathrm{~J}\). what is the mechanical energy of the system? (A) \(34 \mathrm{~J}\) (B) \(45 \mathrm{~J}\) (C) \(48 \mathrm{~J}\) (D) \(49 \mathrm{~J}\)

A force \(\mathrm{F}=\mathrm{kx}\) (where \(\mathrm{k}\) is positive constant) is acting on a particle. Match column-I and column-II, regarding work done in displacing the particle. $$ \begin{array}{|l|l|} \hline \text { Column - i } & \text { Column - ii } \\ \hline \text { (a) From } \mathrm{x}=-4 \text { to } \mathrm{x}=-2 & \text { (P) Positive } \\ \hline \text { (b) From } \mathrm{x}=-2 \text { to } \mathrm{x}=-4 & \text { (Q) zero } \\ \hline \text { (c) From } \mathrm{x}=-2 \text { to } \mathrm{x}=+2 & \text { (R) negative } \\ \hline \end{array} $$ (A) \(\mathrm{a}-\mathrm{R}, \mathrm{b}-\mathrm{P}, \mathrm{c}-\mathrm{Q}\) (B) \(a-P, b-Q, c-R\) (C) \(a-R, b-Q, c-P\) (D) \(\mathrm{a}-\mathrm{Q}, \mathrm{b}-\mathrm{P}, \mathrm{c}-\mathrm{R}\)

The coefficient of restitution e for a perfectly elastic collision is (A) 1 (B) 0 (C) \(\infty\) (D) \(-1\)

What is the velocity of the bob of a simple pendulum at its mean position, if it is able to rise to vertical height of \(18 \mathrm{~cm}\) (Take \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\) ) (A) \(0.4 \mathrm{~m} / \mathrm{s}\) (B) \(4 \mathrm{~m} / \mathrm{s}\) (C) \(1.8 \mathrm{~m} / \mathrm{s}\) (D) \(0.6 \mathrm{~m} / \mathrm{s}\)
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