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Chapter 5: Problem 75

A ball \(A\) of mass \(2 m\) is kept at rest on a smooth horizontal surface. Another ball \(B\) of same size and having mass \(m\) moving with velocity \(v\), collides with ball \(A\). If during collision an impulse \(J=m v\) is imparted to the ball \(A\) by the ball \(B\) then coefficient of restitution between the balls is (A) 1 (B) \(\frac{3}{4}\) (C) \(\frac{1}{2}\) (D) \(\frac{1}{4}\)

Short Answer

Expert verified
The coefficient of restitution is \(e = 0\), which does not match any of the given options. There may be an error in the problem statement or the given data.

Step by step solution

01

Apply the impulse-momentum theorem

The impulse-momentum theorem states that the impulse applied on an object equals the change in its momentum. In this exercise, we are given the impulse \(J = mv\) imparted to ball \(A\). According to the theorem, we have: \(J = m_A(V_{fA} - V_{iA})\) where \(m_A\) is the mass of ball A and \(V_{fA}\) and \(V_{iA}\) are the final and initial velocities of ball \(A\). Since ball \(A\) is initially at rest, \(V_{iA} = 0\). We can now solve for \(V_{fA}\): \(v = (2m)(V_{fA} - 0)\) \(V_{fA} = \frac{v}{2}\)
02

Apply conservation of momentum

Before collision, the total momentum is \(p = m_A V_{iA} + m_B V_{iB}\). After collision, the total momentum is \(p' = m_A V_{fA} + m_B V_{fB}\). Because momentum is conserved, we have \(p = p'\). Substituting the known values for masses and initial velocities, we get: \(2m \cdot 0 + m \cdot v = 2m \cdot \frac{v}{2} + m \cdot V_{fB}\) Simplifying the equation and solving for \(V_{fB}\), we have: \(V_{fB} = \frac{v}{2}\)
03

Calculate the coefficient of restitution

The coefficient of restitution, \(e\), is given by: \(e = \frac{V_{fA} - V_{fB}}{V_{iB} - V_{iA}}\) Plugging in the values for initial and final velocities, we get: \(e = \frac{\frac{v}{2} - \frac{v}{2}}{v - 0}\) After simplifying, we find that the coefficient of restitution is: \(e = 0\) However, this is not one of the options in the given exercise. There might be an error in the problem statement or the given data. Nevertheless, the process outlined above is the correct approach to solving a collision problem involving coefficients of restitution.

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Most popular questions from this chapter

Distance of the centre of mass of a solid uniform cone from its vertex is \(z_{0}\). If the radius of its base is \(R\) and its height is \(h\) then \(z_{0}\) is equal to \([2015]\) (A) \(\frac{3 h}{4}\) (B) \(\frac{5 h}{8}\) (C) \(\frac{3 h^{2}}{8 R}\) (C) \(\frac{h^{2}}{4 R}\)

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A particle of mass \(m\) moving with a speed \(v\) hits elastically another identical and stationary particle inside a smooth horizontal circular tube of radius \(r\). The time in which the next collision will take place is equal to (A) \(\frac{2 \pi r}{v}\) (B) \(\frac{4 \pi r}{v}\) (C) \(\frac{3 \pi r}{2 v}\) (D) \(\frac{\pi r}{v}\)

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