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Chapter 8: Problem 2

A 16 -g mass is moving in the \(+x\) -direction at \(30 \mathrm{~cm} / \mathrm{s}\), while a 4.0-g mass is moving in the \(-x\) -direction at \(50 \mathrm{~cm} / \mathrm{s}\). They collide head on and stick together. Find their velocity after the collision. Assume negligible friction. This is a completely inelastic collision for which \(\mathrm{KE}\) is not conserved, although momentum is. Let the \(16-\mathrm{g}\) mass be \(m_{1}\) and the \(4.0\) -g mass be \(m_{2}\). Take the \(+x\) -direction to be positive. That means that the velocity of the 4.0-g mass has a scalar value of \(v_{2 x}=-50 \mathrm{~cm} / \mathrm{s}\). We apply the law of conservation of momentum to the system consisting of the two masses: Momentum before impact = Momentum after impact $$ \begin{aligned} m_{1} v_{1 x}+m_{2} v_{2 x} &=\left(m_{1}+m_{2}\right) v_{x} \\ (0.016 \mathrm{~kg})(0.30 \mathrm{~m} / \mathrm{s})+(0.0040 \mathrm{~kg})(-0.50 \mathrm{~m} / \mathrm{s}) &=(0.020 \mathrm{~kg}) v_{x} \\ v_{x} &=+0.14 \mathrm{~m} / \mathrm{s} \end{aligned} $$ (Notice that the 4.0-g mass has negative momentum.) Hence, \(\overrightarrow{\mathbf{v}}=0.14 \mathrm{~m} / \mathrm{s}\) - POSITIVE \(X\) -DIRECTION

Short Answer

Expert verified
The final velocity is 0.14 m/s in the positive x-direction.

Step by step solution

01

Define the Problem

We have two masses: one with mass 16 g moving at 30 cm/s in the +x direction and another with mass 4 g moving at 50 cm/s in the -x direction. They will collide and stick together, making this collision inelastic.
02

Convert Units to SI

Convert the given masses to kilograms: \(m_1 = 16 \text{ g} = 0.016 \text{ kg}\) and \(m_2 = 4 \text{ g} = 0.004 \text{ kg}\). Convert the velocities to meters per second: \(v_{1x} = 30 \text{ cm/s} = 0.30 \text{ m/s}\) and \(v_{2x} = -50 \text{ cm/s} = -0.50 \text{ m/s}\).
03

Apply Conservation of Momentum

Use the law of conservation of momentum which states that the total momentum before collision equals the total momentum after collision: \[ m_1 v_{1x} + m_2 v_{2x} = (m_1 + m_2)v_x \]
04

Plug in Values and Solve for Final Velocity

Substitute the known values into the momentum equation: \[(0.016 \text{ kg})(0.30 \text{ m/s}) + (0.004 \text{ kg})(-0.50 \text{ m/s}) = (0.020 \text{ kg})v_x\]Simplify and solve for \(v_x\): \[0.0048 \text{ kg m/s} - 0.0020 \text{ kg m/s} = 0.020 \text{ kg} \times v_x\]\[0.0028 \text{ kg m/s} = 0.020 \text{ kg} \times v_x\]\[v_x = \frac{0.0028 \text{ kg m/s}}{0.020 \text{ kg}} = 0.14 \text{ m/s}\]
05

Determine Direction of Velocity

Since the result \(v_x = 0.14 \text{ m/s}\) is positive, the velocity after the collision is in the positive x-direction.

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Key Concepts

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

Inelastic Collision
An inelastic collision is a type of collision where the colliding objects stick together after impact. Unlike elastic collisions where objects bounce off and move separately, inelastic collisions tend to result in a single, combined mass post-collision. In our exercise, we see the two objects behave this way after they collide. They travel together with a shared velocity. The important thing to remember about inelastic collisions is that while momentum is conserved, kinetic energy is not. This means some of the initial kinetic energy is transformed into other forms of energy, such as heat or sound, resulting in the objects sticking together.
Kinetic Energy
Kinetic energy, denoted as KE, is the energy possessed by an object due to its motion. It's calculated using the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the velocity of the object. In our context, the initial kinetic energy of the two colliding masses was not fully conserved during the collision. When they collided inelastically, some kinetic energy was transformed into other types, such as sound or heat. This is typical in inelastic collisions and is part of why objects stick together. It鈥檚 crucial to differentiate between momentum and kinetic energy here, as only momentum is conserved during these types of collisions.
Unit Conversion
Unit conversion is often a critical step in physics problems to ensure consistency and correctness of results. In the original problem, masses were given in grams and velocities in centimeters per second. To make calculations easier and more standard, we need to convert these to SI units.
  • Mass conversion: 16 g becomes 0.016 kg and 4 g becomes 0.004 kg by dividing by 1000 since 1 kg = 1000 g.
  • Velocity conversion: 30 cm/s becomes 0.30 m/s and -50 cm/s becomes -0.50 m/s by dividing by 100 since 1 m = 100 cm.
Performing these conversions at the start helps in streamlining further calculations and in preventing errors that can arise from using inconsistent units.
Velocity Calculation
Velocity calculation is at the heart of solving physics problems involving collisions. In our exercise, the velocities of the masses before the collision were crucial for applying the conservation of momentum theorem. The final velocity of the combined mass post-collision can be found using:\[ m_1 v_{1x} + m_2 v_{2x} = (m_1 + m_2) v_x \]Substituting the masses and velocities after converting them, we find:\( v_x = \frac{0.0028}{0.020} = 0.14 \) m/s.This final velocity indicates that after collision, the new combined mass moves in the positive x-direction with a velocity of 0.14 m/s. This directionality is important as it reflects the combined effect of initial momentums and shows how factors like object mass and initial velocity influence the result.

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Most popular questions from this chapter

A 6000 -kg truck traveling north at \(5.0 \mathrm{~m} / \mathrm{s}\) collides with a \(4000-\mathrm{kg}\) truck moving west at \(15 \mathrm{~m} / \mathrm{s}\). If the two trucks remain locked together after impact, with what speed and in what direction do they move immediately after the collision?

Two identical balls undergo a collision at the origin of coordinates. Before collision their scalar velocity components are \(\left(u_{x}=40 \mathrm{~cm} / \mathrm{s}, u_{y}=0\right)\) and \(\left(u_{x}=-30 \mathrm{~cm} / \mathrm{s}, u_{y}=20 \mathrm{~cm} / \mathrm{s}\right) .\) After collision, the first ball (the one moving along the \(x\) -axis) is standing still. Find the scalar velocity components of the second ball. [Hint: After the collision, the moving ball must have all of the momentum of the system.]

A pendulum consisting of a ball of mass \(m\) is released from the position shown in Fig. \(8-2\) and strikes a block of mass \(M\). The block slides a distance \(D\) before stopping under the action of a steady friction force of \(0.20 M g .\) Find \(D\) if the ball rebounds to an angle of \(20^{\circ}\). The pendulum ball falls through a height \(\left(L-L \cos 37^{\circ}\right)=0.201 L\) and rebounds to a height \(\left(L-L \cos 20^{\circ}\right)=\) \(0.0603 L\). Because \((m g h)_{\text {top }}=\left(\frac{1}{2} m v^{2}\right)_{\text {bottom }}\) for the ball, its speed at the bottom is \(v=\sqrt{2 g h}\). Thus, just before it hits the block, the ball has a speed equal to \(\sqrt{2 g(0.201 L)}\). Since the ball rises up to a height of \(0.0603 L\) after the collision, it must have rebounded with an initial speed of \(\sqrt{2 g(0.0603 L)}\). Although \(\mathrm{KE}\) is not conserved in the collision, momentum is. Therefore, for the collision, Momentum just before \(=\) Momentum just after $$ m \sqrt{2 g(0.201 L)}+0=-m \sqrt{2 g(0.0603 L)}+M V $$ where \(V\) is the velocity of the block just after the collision. (Notice the minus sign on the momentum of the rebounding ball.) Solving this equation, we find $$ V=\frac{m}{M} 0.981 \sqrt{g L} $$ The block uses up its translational KE doing work against friction as it slides a distance \(D\). Therefore, $$ \frac{1}{2} M V^{2}=F_{\mathrm{f}} D \quad \text { or } \quad \frac{1}{2} M(0.963 g L)\left(\frac{m}{M}\right)^{2}=(0.2 M g)(D) $$ from which \(D=2.4(m / M)^{2} L\).

A \(15-\mathrm{g}\) bullet moving at \(300 \mathrm{~m} / \mathrm{s}\) passes through a \(2.0-\mathrm{cm}\) -thick sheet of foam plastic and emerges with a speed of \(90 \mathrm{~m} / \mathrm{s}\). Assuming that the speed change takes place uniformly, what average force impeded the bullet's motion through the plastic? We can determine the change in momentum, and that suggests using the impulse equation to find the force \(F\) on the bullet as it takes a time \(\Delta t\) to pass through the plastic. Taking the initial direction of motion to be positive, $$ F \Delta t=m v_{f}-m v_{i} $$ We can find \(\Delta t\) by assuming uniform deceleration and using \(x=v_{a v} t\), where \(x=0.020 \mathrm{~m}\) and \(v_{a v}=\frac{1}{2}\left(v_{i}+v_{f}\right)=\) \(195 \mathrm{~m} / \mathrm{s}\). This gives \(\Delta t=1.026 \times 10^{-4} \mathrm{~s}\). Then $$ (F)\left(1.026 \times 10^{-4} \mathrm{~s}\right)=(0.015 \mathrm{~kg})(90 \mathrm{~m} / \mathrm{s})-(0.015 \mathrm{~kg})(300 \mathrm{~m} / \mathrm{s}) $$ which yields \(F=-3.1 \times 10^{4} \mathrm{~N}\) as the average retarding force. How could this problem have been solved using \(F=m a\) instead of the impulse equation? By using energy methods?

Two girls (masses \(m_{1}\) and \(m_{2}\) ) are on roller skates and stand at rest, close to each other and face to face. Girl- 1 pushes squarely against girl- 2 and sends her moving backward. Assuming the girls move freely on their skates, write an expression for the speed with which girl- 1 moves. We take the two girls to comprise the system under consideration. The problem states that girl- 2 moves "backward," so let that be the negative direction; therefore, the "forward" direction is positive. There is no resultant external force on the system (the push of one girl on the other is an internal force), and so momentum is conserved: $$ \begin{aligned} \text { Momentum before } &=\text { Momentum after } \\ 0 &=m_{1} v_{1}+m_{2} v_{2} \end{aligned} $$ from which \(\quad v_{1}=-\frac{m_{2}}{m_{1}} v_{2}\) Girl- 1 recoils with this speed. Notice that if \(m_{2} / m_{1}\) is very large, \(v_{1}\) is much larger than \(v_{2}\). The velocity of girl- \(1, \overrightarrow{\mathbf{v}}_{1}\), points in the positive forward direction. The velocity of girl- \(2, \overrightarrow{\mathbf{v}}_{2}\), points in the negative backward direction. If we put numbers into the equation, \(v_{2}\) would have to be negative and \(v_{1}\) would come out positive.
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