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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

If the water falls from a dam into a turbine wheel \(19.6 \mathrm{~m}\) below, then the velocity of water at the turbine is \(\ldots \ldots\) \(\left(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) \(9.8 \mathrm{~m} / \mathrm{s}\) (B) \(19.6 \mathrm{~m} / \mathrm{s}\) (C) \(39.2 \mathrm{~m} / \mathrm{s}\) (D) \(98.0 \mathrm{~m} / \mathrm{s}\)

A metal ball of mass \(2 \mathrm{~kg}\) moving with a velocity of $36 \mathrm{~km} / \mathrm{h}$ has a head on collision with a stationary ball of mass \(3 \mathrm{~kg}\). If after the collision, the two balls move together, the loss in kinetic energy due to collision is (A) \(40 \mathrm{~J}\) (B) \(60 \mathrm{~J}\) (C) \(100 \mathrm{~J}\) (D) \(140 \mathrm{~J}\)

If a man increase his speed by \(2 \mathrm{~m} / \mathrm{s}\), his $\mathrm{K} . \mathrm{E}$. is doubled, the original speed of the man is (A) \((2+2 \sqrt{2}) \mathrm{m} / \mathrm{s}\) (B) \((2+\sqrt{2}) \mathrm{m} / \mathrm{s}\) (C) \(4 \mathrm{~m} / \mathrm{s}\) (D) \((1+2 \sqrt{2}) \mathrm{m} / \mathrm{s}\)

A mass of \(\mathrm{M} \mathrm{kg}\) is suspended by a weight-less string, the horizontal force that is required to displace it until the string makes an angle of \(60^{\circ}\) with the initial vertical direction is (A) \(\mathrm{Mg} / \sqrt{3}\) (B) \(\mathrm{Mg} \cdot \sqrt{2}\) (C) \(\mathrm{Mg} / \sqrt{2}\) (D) \(\mathrm{Mg} \cdot \sqrt{3}\)

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) \(24 \mathrm{~J}\)
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