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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 (θ2) 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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