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Chapter 7: Problem 103

A particle \(\left(M_{1}=1.00 \mathrm{~kg}\right)\) moving at \(30.0^{\circ}\) downward from the horizontal with \(v_{1}=2.50 \mathrm{~m} / \mathrm{s}\) hits a second particle \(\left(M_{2}=2.00 \mathrm{~kg}\right),\) which was at rest momentarily. After the collision, the speed of \(M_{1}\) was reduced to \(.500 \mathrm{~m} / \mathrm{s}\), and it was moving at an angle of \(32^{\circ}\) downward with respect to the horizontal. Assume the collision is elastic. a) What is the speed of \(M_{2}\) after the collision? b) What is the angle between the velocity vectors of \(M_{1}\) and \(M_{2}\) after the collision?

Short Answer

Expert verified
Question: In an elastic collision between two particles M1 and M2, particle M1 initially moves at 2.50 m/s at an angle of 30 degrees downward from the horizontal and particle M2 is at rest. After the collision, particle M1 moves at 1.75 m/s at an angle of 32 degrees upward from the horizontal. Determine the speed and angle of particle M2 after the collision and the angle between the velocity vectors of the particles after the collision. Answer: Follow the step-by-step solution to find the speed (v2) and angle (胃2) of particle M2 after the collision, as well as the angle between the velocity vectors of particles M1 and M2.

Step by step solution

01

Determine the initial momentum of each particle

To have a clear view of the collision, sketch a diagram with the velocities of both particles. For particle M1, we have the initial velocity given to be v1 = 2.50 m/s at an angle 30 degrees downward from the horizontal, while particle M2 is at rest, so its initial velocity is 0 m/s. Now, determine the initial momentum of both particles in x and y coordinates: $$p_{1x} = m_{1}v_{1}\cos(30^{\circ})$$ $$p_{1y} = -m_{1}v_{1}\sin(30^{\circ})$$ $$p_{2x} = 0$$ $$p_{2y} = 0$$
02

Determine the final momentum of the particles

We also need the final momentum of both particles M1 and M2 after the collision. To do this, we must use the final velocities and angles given for the particle M1 and unknowns for the particle M2. We want to find out \(v_{2}, \theta_{2}\), first determine the final momentum of M1 after the collision: $$p'_{1x} = m_{1}v'_{1}\cos(32^{\circ})$$ $$p'_{1y} = -m_{1}v'_{1}\sin(32^{\circ})$$ For M2, we have unknowns as the magnitude of the velocity, \(v_{2}\) and angle \(\theta_{2}\): $$p'_{2x} = m_{2}v_{2}\cos(\theta_{2})$$ $$p'_{2y} = -m_{2}v_{2}\sin(\theta_{2})$$
03

Applying conservation of momentum

Applying momentum conservation along x and y directions respectively, we get: $$p_{1x} + p_{2x} = p'_{1x} + p'_{2x}$$ $$p_{1y} + p_{2y} = p'_{1y} + p'_{2y}$$ Substituting the expressions for momenta, we get $$m_{1}v_{1}\cos(30^{\circ}) = m_{1}v'_{1}\cos(32^{\circ}) + m_{2}v_{2}\cos(\theta_{2}) \qquad (1)$$ $$-m_{1}v_{1}\sin(30^{\circ}) = -m_{1}v'_{1}\sin(32^{\circ}) - m_{2}v_{2}\sin(\theta_{2}) \qquad (2)$$
04

Applying conservation of kinetic energy

For the collision to be elastic, the kinetic energy should also be conserved. So, $$\frac{1}{2}m_{1}v_{1}^2 + \frac{1}{2}m_{2}0^2 = \frac{1}{2}m_{1}v'_{1}^2 + \frac{1}{2}m_{2}v_{2}^2$$ Solving for \(v_{2}\), we get $$v_{2} = \sqrt{\frac{(2m_{1}v_{1}^2 - 2m_{1}v'_{1}^2)}{2m_{2}}} \qquad (3)$$
05

Solving for \(v_{2}\) and \(\theta_{2}\)

Using equation (3), we can calculate the value of \(v_{2}\). Now, knowing the value for \(v_{2}\), we can solve for the angle \(\theta_{2}\) using equations (1) and (2) simultaneously. Substitute the calculated value of \(v_{2}\) into equation (1) and (2), and then solve for \(\theta_{2}\).
06

Finding the angle between velocity vectors of particles after collision

Finally, we can find the angle between the velocity vectors of M1 and M2 after the collision by using trigonometry. $$\cos^{-1}(\cos\theta_{2}\cos(32^{\circ}) + \sin\theta_{2}\sin(32^{\circ}))$$

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Key Concepts

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

Conservation of Momentum
The principle of conservation of momentum is crucial in understanding elastic collisions. Momentum, a product of mass and velocity, is a vector quantity, meaning it has both magnitude and direction. In any collision, the total momentum before and after the event stays the same if no external forces act on the system.

In the given problem, particle M1 has an initial momentum because it is in motion, while particle M2 is at rest, contributing no initial momentum. After the collision, both particles have momentum in the x and y directions. The conservation of momentum is applied in two directions:
  • X-direction: The horizontal components of momentum before and after the collision are equal.
  • Y-direction: The vertical components of the momentum also balance out before and after the collision.
This principle helps us solve for unknowns by setting up equations based on the initial and final momentum components of both particles. The formulas combine these momentums, taking into account the angles and the final velocities, which are crucial for finding the missing parts in this puzzle of motion.
Conservation of Kinetic Energy
In an elastic collision, like the one in this exercise, the conservation of kinetic energy is another key principle. Kinetic energy is given by the formula \( \frac{1}{2}mv^2 \), which means that for any particle with mass \( m \) and velocity \( v \), its kinetic energy is dependent solely on these two parameters.

For the collision to be fully elastic, the total kinetic energy of the system must remain constant before and after the event. This doesn't mean that each object retains the same energy, but that their combined energy does. The given kinetic energy formula applies before and after the collision:
  • Before the collision, M1 possesses kinetic energy with its initial velocity, and M2 has no energy as it is at rest.
  • After the collision, both M1 and M2 will have kinetic energy as they move away at different speeds.
This principle allows us to set up an equation for the kinetic energies before and after to find unknown values like the speed of M2 following the impact. It ensures that despite the alteration in speed and direction, the energy is merely transferred, not lost.
Angle Between Velocity Vectors
Finding the angle between velocity vectors of two particles after a collision provides insight into their trajectories. This calculation considers the directions in which the particles move post-collision, which can show if the collision is more head-on or glancing.

In this problem, after solving for the velocity vector of M2 using conservation laws, we can then find the angle between M1 and M2's velocity vectors. The angle is derived from geometric considerations of vectors. We use cosine law or trigonometric identities to relate angles and directions:
  • Dot Product: The angle between two vectors can be found using the dot product formula, which involves calculating how aligned the vectors are.
  • Given Angles: Using known angles such as M1's post-collision angle, we can use trigonometric relationships to deduce the angle between the two vectors.
  • Vector Components: By considering the x and y components of each velocity, the calculations can be broken down into manageable parts.
This angle is crucial for answering part b of the exercise, giving a comprehensive view of the situational dynamics brought by the collision.

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Most popular questions from this chapter

A ball with mass \(m=0.210 \mathrm{~kg}\) and kinetic energy \(K_{1}=2.97 \mathrm{~J}\) collides elastically with a second ball of the same mass that is initially at rest. \(m_{1}\) After the collision, the first ball moves away at an angle of \(\theta_{1}=\) \(30.6^{\circ}\) with respect to the horizontal, as shown in the figure. What is the kinetic energy of the first ball after the collision?

A fireworks projectile is launched upward at an angle above a large flat plane. When the projectile reaches the top of its flight, at a height of \(h\) above a point that is a horizontal distance \(D\) from where it was launched, the projectile explodes into two equal pieces. One piece reverses its velocity and travels directly back to the launch point. How far from the launch point does the other piece land? a) \(D\) b) \(2 D\) c) \(3 D\) d) \(4 D\)

A hockey puck with mass \(0.250 \mathrm{~kg}\) traveling along the blue line (a blue-colored straight line on the ice in a hockey rink) at \(1.50 \mathrm{~m} / \mathrm{s}\) strikes a stationary puck with the same mass. The first puck exits the collision in a direction that is \(30.0^{\circ}\) away from the blue line at a speed of \(0.750 \mathrm{~m} / \mathrm{s}\) (see the figure). What is the direction and magnitude of the velocity of the second puck after the collision? Is this an elastic collision?

Three birds are flying in a compact formation. The first bird, with a mass of \(100 . \mathrm{g}\) is flying \(35.0^{\circ}\) east of north at a speed of \(8.00 \mathrm{~m} / \mathrm{s}\). The second bird, with a mass of \(123 \mathrm{~g}\), is flying \(2.00^{\circ}\) east of north at a speed of \(11.0 \mathrm{~m} / \mathrm{s}\). The third bird, with a mass of \(112 \mathrm{~g}\), is flying \(22.0^{\circ}\) west of north at a speed of \(10.0 \mathrm{~m} / \mathrm{s}\). What is the momentum vector of the formation? What would be the speed and direction of a \(115-\mathrm{g}\) bird with the same momentum?

A golf ball of mass \(45.0 \mathrm{~g}\) moving at a speed of \(120 . \mathrm{km} / \mathrm{h}\) collides head on with a French TGV high-speed train of mass \(3.8 \cdot 10^{5} \mathrm{~kg}\) that is traveling at \(300 . \mathrm{km} / \mathrm{h}\). Assuming that the collision is elastic, what is the speed of the golf ball after the collision? (Do not try to conduct this experiment!)
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