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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

Potential Energy Function A single conservative force \(F(x)\) acts on a \(1.0 \mathrm{~kg}\) particle that moves along an \(x\) axis. The potential energy \(U(x)\) associated with \(F(x)\) is given by $$ U(x)=(-4.00 \mathrm{~J} / \mathrm{m}) e^{(-x /(4.00 \mathrm{~m}))} $$ At \(x=5.0 \mathrm{~m}\) the particle has a kinetic energy of \(2.0 \mathrm{~J}\). (a) What is the mechanical energy of the system? (b) Make a plot of \(U(x)\) as a function of \(x\) for \(0 \leq x \leq 10 \mathrm{~m}\), and on the same graph draw the line that represents the mechanical energy of the system. Use part (b) to determine (c) the least value of \(x\) and (d) the greatest value of \(x\) between which the particle can move. Use part (b) to determine (e) the maximum kinetic energy of the particle and (f) the value of \(x\) at which it occurs. (g) Determine the equation for \(F(x)\) as a function of \(x\). (h) For what (finite) value of \(x\) does \(F(x)=0 ?\)

Worker Pushes Block A worker pushed a \(27 \mathrm{~kg}\) block \(9.2 \mathrm{~m}\) along a level floor at constant speed with a force directed \(32^{\circ}\) below the horizontal. If the coefficient of kinetic friction between block and floor was \(0.20\), what were (a) the work done by the worker's force and (b) the increase in thermal energy of the block-floor system?

Rigid Rod A rigid rod of length \(L\) and negligible mass has a ball with mass \(m\) attached to one end and its other end fixed, to form a pendulum. The pendulum is inverted, with the rod straight up, and then released. At the lowest point, what are (a) the ball's speed and (b) the tension in the rod? (c) The pendulum is next released at rest from a horizontal position. At what angle from the vertical does the tension in the rod equal the weight of the ball?

Diatomic Molecule The potential energy of a diatomic molecule (a two-atom system like \(\mathrm{H}_{2}\) or \(\mathrm{O}_{2}\) ) is given by $$ U=\frac{A}{r^{12}}-\frac{B}{r^{6}} $$ where \(r\) is the separation of the two atoms of the molecule and \(A\) and \(B\) are positive constants. This potential energy is associated with the force that binds the two atoms together. (a) Find the equilibrium separation-that is, the distance between the atoms at which the force on each atom is zero. Is the force repulsive (the atoms are pushed apart) or attractive (they are pulled together) if their separation is (b) smaller and (c) larger than the equilibrium separation?

Two Blocks and a Spring A block of mass \(m_{A}=2.0 \mathrm{~kg}\) slides along a frictionless table with a speed of \(10 \mathrm{~m} / \mathrm{s}\). Directly in front of it, and moving in the same direction, is a block of mass \(m_{B}=5.0 \mathrm{~kg}\) moving at \(3.0 \mathrm{~m} / \mathrm{s}\). A massless spring with spring constant \(k=1120\) \(\mathrm{N} / \mathrm{m}\) is attached to the near side of \(m_{B}\), as shown in Fig. \(10-57\). When the blocks collide, what is the maximum compression of the spring? (Hint: At the moment of maximum compression of the spring, the two blocks move as one. Find the velocity by noting that the collision is completely inelastic at this point.)
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