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Chapter 10: Problem 94

Frictionless Ramp In Fig. \(10-64\), block \(A\) of mass \(m_{A}\) slides from rest along a frictionless ramp from a height of \(2.50 \mathrm{~m}\) and then collides with stationary block \(B\), which has mass \(m_{B}=2.00 m_{A} .\) After the collision, block \(B\) slides into a region where the coefficient of kinetic friction is \(0.500\) and comes to a stop in distance \(d\) within that region. What is the value of distance \(d\) if the collision is (a) elastic and (b) completely inelastic?

Short Answer

Expert verified
Elastic collision: \( d = 7.35\mathrm{~m} \)Completely inelastic collision: \( d = 1.67\mathrm{~m} \)

Step by step solution

01

Determine the speed of block A at the bottom of the ramp

Use the conservation of mechanical energy to find the speed of block A as it reaches the bottom of the ramp. Initial potential energy = final kinetic energy: \[ m_A g h = \frac{1}{2} m_A v_A^2 \]\[ v_A = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 2.5} \]\[ v_A = 7 \mathrm{~m/s} \]
02

Calculate the velocity of both blocks after an elastic collision

Use the conservation of momentum and kinetic energy for an elastic collision. Let the velocities after the collision be \(v'_A\) for block A and \(v'_B\) for block B: \[ m_A v_A = m_A v'_A + m_B v'_B \] \[ \frac{1}{2} m_A v_A^2 = \frac{1}{2} m_A v'_A^2 + \frac{1}{2} m_B v'_B^2 \] Substituting values: \[ m_A \times 7 = m_A v'_A + 2m_A v'_B \] \[ \frac{1}{2} m_A \times 7^2 = \frac{1}{2} m_A v'_A^2 + \frac{1}{2} \times 2m_A v'_B^2 \]Solve system of equations to get: \[ v'_A = -1\mathrm{~m/s} \] and \[ v'_B = 6\mathrm{~m/s} \]
03

Calculate the velocity of both blocks after a completely inelastic collision

In a completely inelastic collision, both blocks stick together. Use the conservation of momentum: \[ m_A v_A = (m_A + m_B) v_f \] \[ v_f = \frac{m_A \times 7}{m_A + 2m_A} = \frac{7}{3} \mathrm{~m/s}\]
04

Determine the distance block B travels with friction (elastic case)

Kinetic energy of block B is converted to work done against friction: \[ \frac{1}{2} m_B v'_B^2 = \mu m_B g d \] \[ \frac{1}{2} \times 2 m_A \times 6^2 = 0.5 \times 2 m_A \times 9.8 \times d \] \[ d = \frac{36}{4.9} = 7.35 \mathrm{~m} \]
05

Determine the distance block B travels with friction (completely inelastic case)

Use the initial speed \(v_f\) found earlier for the combined masses: \[ \frac{1}{2} (m_A + m_B) v_f^2 = \mu (m_A + m_B) g d \] \[ \frac{1}{2} \times 3m_A \times \left(\frac{7}{3}\right)^2 = 0.5 \times 3 m_A \times 9.8 \times d \] \[ d = \frac{49}{29.4} = 1.67 \mathrm{~m} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conservation of Energy
The principle of Conservation of Energy is critical for solving ramp problems. It states that the total energy in an isolated system remains constant. For a frictionless ramp, potential energy at the top is converted into kinetic energy at the bottom. Here, block A starts from rest at a height of 2.5 meters. Its initial potential energy is transformed into kinetic energy as it descends. Using the equation \[ m_A g h = \frac{1}{2} m_A v_A^2 \], we find block A鈥檚 velocity at the bottom to be 7 m/s.
Elastic Collision
An elastic collision is one where both momentum and kinetic energy are conserved. For block A and block B: right before and after collision, the sum of their momenta and kinetic energies remain the same. By applying the conservation of momentum and kinetic energy formulas \[ m_A v_A = m_A v'_A + m_B v'_B \] and \[ \frac{1}{2} m_A v_A^2 = \frac{1}{2} m_A v'_A^2 + \frac{1}{2} m_B v'_B^2 \], we calculate post-collision velocities: block A has -1 m/s and block B has 6 m/s.
Inelastic Collision
In an inelastic collision, only momentum is conserved, not kinetic energy. A completely inelastic collision means the objects stick together post-collision. We use the momentum conservation equation \[ m_A v_A = (m_A + m_B) v_f \]. Substituting values, we find the combined velocity to be \[ v_f = \frac{7}{3} \mathrm{~m/s} \]. Notice the distinct change: kinetic energy loss is represented as heat or sound.
Kinetic Friction
Kinetic friction acts to stop a moving object. For block B, encountering friction part, its kinetic energy converts to work done against friction: \[ \frac{1}{2} m_B v'^2 = \mu m_B g d \]. In the elastic collision, block B's initial speed is 6 m/s. Solving for distance, we get \[ d = 7.35 \mathrm{~m} \]. In the inelastic case, where both blocks move together at \[ \frac{7}{3} \mathrm{~m/s} \], distance reduces to \[ d = 1.67 \mathrm{~m} \]. Notice the effect of friction in energy dissipation.
Kinematics
Kinematics focuses on the motion specifics of objects without considering the forces causing it. For block A's descent along the ramp, kinematic equations helped find the velocity from height: \[ v = \sqrt{2gh} \]. Post-collision, analyzing block B鈥檚 travel considers initial velocity and deceleration due to friction. Such breakdown aids visualization of step-wise energy and momentum transitions, enhancing overall understanding.

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Most popular questions from this chapter

Bullet Hits Wall A \(30 \mathrm{~g}\) bullet, with a horizontal velocity of \(500 \mathrm{~m} / \mathrm{s}\), comes to a stop \(12 \mathrm{~cm}\) within a solid wall. (a) What is the change in its mechanical energy? (b) What is the magnitude of the average force from the wall stopping it?

Closing the Door A student is in her dorm room, sitting on her bed doing her physics homework. The door to her room is open. All of a sudden, she hears the voice of her ex-boyfriend talking to the girl in the room next door. She wants to shut the door quickly, so she throws a superball (which she keeps next to her bed for this purpose) against the door. The ball follows the path shown in Fig. 10-69. It hits the door squarely and bounces straight back. (a) If the ball has a mass \(m\), hits the door with a speed \(v\), and bounces back with a speed equal to \(v\), what is the change in the ball's momentum? (b) If the ball was in contact with the door for a time \(\Delta t\), what was the average force that the door exerted on the ball? (c) Would she have been better off with a clay ball of the same mass that stuck to the door? Explain your reasoning.

You drop a \(2.00\) kg textbook to a friend who stands on the ground \(10.0 \mathrm{~m}\) below the textbook with outstretched hands \(1.50 \mathrm{~m}\) above the ground (Fig. \(10-26\) ). (a) How much work \(W^{\text {grav }}\) is done on the textbook by the gravitational force as it drops to your friend's hands? (b) What is the change \(\Delta U\) in the gravitational potential energy of the textbook-Earth system during the drop? If the gravitational potential energy \(U\) of that system is taken to be zero at ground level, what is \(U\) when the textbook (c) is released and (d) reaches the hands? Now take \(U\) to be \(100 \mathrm{~J}\) at ground level and again find (e) \(W\) grav (f) \(\Delta U,(\mathrm{~g}) U\) at the release point, and (h) \(U\) at the hands.

What is the spring constant of a spring that stores \(25 \mathrm{~J}\) of elastic potential energy when compressed by \(7.5 \mathrm{~cm}\) from its relaxed length?

Three Balls ball \(A\) with an initial speed of \(10 \mathrm{~m} / \mathrm{s}\) collides elastically with stationary balls \(B\) and \(C\), whose centers are on a line perpendicular to the initial velocity of ball \(A\) and that are initially in contact with each other. The three balls are identical. Ball \(A\) is aimed directly at the contact point, and all motion is frictionless. After the collision, what are the velocities of (a) ball \(B,(\mathrm{~b})\) ball \(C\), and \((\mathrm{c})\) ball \(A ?\) (Hint: With friction absent, each impulse is directed along the line connecting the centers of the colliding balls, normal to the colliding surfaces.)
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