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Chapter 15: Problem 35

The 5 -Mg bus \(B\) is traveling to the right at \(20 \mathrm{~m} / \mathrm{s}\) Meanwhile a \(2-\mathrm{Mg}\) car \(A\) is traveling at \(15 \mathrm{~m} / \mathrm{s}\) to the right. If the vehicles crash and become entangled, determine their common velocity just after the collision. Assume that the vehicles are free to roll during collision.

Short Answer

Expert verified
Recalculating the equation \( (5 Mg \times 20 m/s) +(2 Mg \times 15 m/s) = (5 Mg + 2 Mg) \times V\), the answer is \(17.14 ms^{-1}\). Hence, the common velocity of the vehicles after the collision is approximately \(17.14 m/s\).

Step by step solution

01

Calculate Initial Total Momentum

The initial total momentum of the system is given by the sum of the momenta of the bus and the car before the collision. Momentum is calculated as mass multiplied by velocity. For the bus, this is \(5 Mg \times 20 m/s\) and for the car, this is \(2 Mg \times 15 m/s\). Hence, the initial total momentum equals \( (5 Mg \times 20 m/s) +(2 Mg \times 15 m/s)\).
02

Set Up Conservation of Momentum Equation

The total momentum after the collision will be the combined mass of both the car and bus multiplied by their common velocity, \(V\), which we are trying to find. So this can be written as \( (5 Mg + 2 Mg) \times V\)\.
03

Solve for Common Velocity

Since the total initial momentum (from step 1) must equal the total final momentum (from step 2), we can set these two equal to each other and solve for the common velocity, \(V\). This gives us \( (5 Mg \times 20 m/s) +(2 Mg \times 15 m/s) = (5 Mg + 2 Mg) \times V\). Solving for \(V\) gives us the common velocity of the bus and car after the collision.

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Key Concepts

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

Inelastic Collision
During an inelastic collision, two objects crash into each other and stick together. This implies that they move as a single combined object after the collision. Contrary to elastic collisions, where kinetic energy is conserved, in inelastic collisions, only momentum is conserved. Energy transformation into other forms, like heat or deformation, often occurs.
In the exercise we have on hand, when the 5-Mg bus and 2-Mg car collide, they become entangled. As they stick together, this is a textbook example of an inelastic collision. The focus shifts from merely the speed to how the system behaves as a whole. By considering only the momentum, we can bypass the complexities introduced by various energy conversions post-collision.
Momentum Calculation
Momentum is a fundamental concept in physics utilized to predict the outcome of interactions like collisions. It is given by the product of mass and velocity and is expressed mathematically as: \[ p = m \times v \] where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity. In this exercise, calculating the individual momenta of the bus and car before the collision is necessary.
  • The bus, with a mass of 5 Mg moving at 20 m/s, has a momentum of \( 5 \times 20 = 100 \, \text{Mg m/s} \).
  • The car, with a mass of 2 Mg moving at 15 m/s, has a momentum of \( 2 \times 15 = 30 \, \text{Mg m/s} \).
Adding these values gives us the total momentum of the system just before the collision, which is crucial because we utilize this in our conservation of momentum calculations.
Velocity Determination
To find the common velocity of two entangled objects after an inelastic collision, we apply the principle of conservation of momentum. According to this principle, the total momentum before collision equals the total momentum after collision. This can be represented with the equation: \[ m_1 v_1 + m_2 v_2 = (m_1 + m_2) V \] Given the exercise, the total initial momentum we calculated as \( 100 \, \text{Mg m/s} + 30 \, \text{Mg m/s} \) must equate to the total final momentum \( V \cdot (5 \text{ Mg} + 2 \text{ Mg}) \).
  • Plugging in the known values, our equation becomes \( 130 \, \text{Mg m/s} = 7 V \).
  • Solving for \( V \), the common velocity is determined to be \( \frac{130}{7} \, \text{m/s} \).
This calculation shows the new shared velocity post-collision, wrapping up our problem solution.

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Most popular questions from this chapter

Two smooth billiard balls \(A\) and \(B\) each have a mass of \(200 \mathrm{~g}\). If \(A\) strikes \(B\) with a velocity \(\left(v_{A}\right)_{1}=1.5 \mathrm{~m} / \mathrm{s}\) as shown, determine their final velocities just after collision. Ball \(B\) is originally at rest and the coefficient of restitution is \(e=0.85 .\) Neglect the size of each ball.

Block \(A\) has a mass of \(2 \mathrm{~kg}\) and slides into an open ended box \(B\) with a velocity of \(2 \mathrm{~m} / \mathrm{s}\). If the box \(B\) has a mass of \(3 \mathrm{~kg}\) and rests on top of a plate \(P\) that has a mass of \(3 \mathrm{~kg}\), determine the distance the plate moves after it stops sliding on the floor. Also, how long is it after impact before all motion ceases? The coefficient of kinetic friction between the box and the plate is \(\mu_{k}=0.2,\) and between the plate and the floor \(\mu_{k}^{\prime}=0.1 .\) Also, the coefficient of static friction between the plate and the floor is \(\mu_{s}^{\prime}=0.12\).

The \(6-\mathrm{kg}\) block is falling downward at \(v_{1}=3 \mathrm{~m} / \mathrm{s}\) when it is \(8 \mathrm{~m}\) from the sandy surface. Determine the average impulsive force acting on the block by the sand if the motion of the block is stopped in time \(1.2 \mathrm{~s}\) once the block strikes the sand. Neglect the distance the block dents into the sand and assume the block does not rebound. Neglect the weight of the block during the impact with the sand.

The pile \(P\) has a mass of \(800 \mathrm{~kg}\) and is being driven into loose sand using the \(300-\mathrm{kg}\) hammer \(C\) which is dropped a distance of \(0.5 \mathrm{~m}\) from the top of the pile. Determine the initial speed of the pile just after it is struck by the hammer. The coefficient of restitution between the hammer and the pile is \(e=0.1 .\) Neglect the impulses due to the weights of the pile and hammer and the impulse due to the sand during the impact.

A ballistic pendulum consists of a 4 -kg wooden block originally at rest, \(\theta=0^{\circ} .\) When a \(2-\mathrm{g}\) bullet strikes and becomes embedded in it, it is observed tlat the block swings upward to a maximum angle of \(\theta=6^{\circ} .\) Estimate the initial speed of the bullet.
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