detail as below Use the following Active Figure representing a head-on collision
ID: 2136214 • Letter: D
Question
detail as below
Use the following Active Figure representing a head-on collision between two particles to complete the exercise. Click 'start' to begin the animation. A particle of mass mi = 12 kg with velocity V1i = 15 m/s collides elastically into a second particle of mass m2 = 17 kg that is initially at rest. Determine the velocities f and V2f of the particles after collision. The particle with mass mi approaches the particle with m2- The particles move off separately after the collision. Forces act on the particles only during the collision, and the elastic nature of the collision means that even though some kinetic energy might have been transferred between particles, no kinetic energy was lost. The only forces to act in any way Display in a New Window that matters are the internal forces between the two objects, and there is no external force to consider. Because no external force acts, the collision does not change the total momentum of the system of two particles. This can be stated as mivi,= m2v2f. Also, because the collision is elastic, the total kinetic energy before collision is the same as the total kinetic energy after collision: 1/2m1v12 = 1/2m1v1f2 + 1/2m2v2f2 Having found relations that must be satisfied by the initial and final velocities, we next solve for v1i and v2i in terms of the known quantities mi, m2l vy,, and v2l. Equation (1) can be rewritten as This may then be used to eliminate v^f from Equation (2). m1v1i2 = (m1v1i - m2v2f)2/m1 + m2v2f2 Expanding the numerator in the first term yields m1v1i2=m1v1i2 - 2m1m2v1iv2f / m1+ m1v2f2 which simplifies to become 0 = (m2v2f)(-2m1v1+ m2v2f + miv2f) At least one of the two factors in Equation (6) must be zero. The first factor is zero for v2f = 0 m/s. This would correspond to each particle keeping its initial velocity, contrary to the assumption that they collide. We discard this solution. Now the other factor in Equation (6) must be zero, which implies v2f = 2m1/m1 + m2 v1i. We may then substitute Equation (7) into Equation (1), m1v1i = m1v1f + m2 2m1/m1 + m2 v1i. Calculating from Equation (7) and Equation (9) gives v2f = 2(12 kg)/(12 kg) + (17 kg)/(15 m/s)= m/s, v1f = 12kg - 17 kg/12 kg + 17 kg (15 m/s) = m/s What result does this predict when the two particles have equal mass and the second one is initially stationary? How could you have predicted that without solving any equations? What if the first mass is large and collides elastically with the second mass? What simple relations are implied between the two final velocities and the initial velocity?Explanation / Answer
V1f=(12-17)*15/(12+17)=-85/29=-2.931 m/s
v2f=2*12*15/(12+17)=12.414 m/s
summary::
1) the velocity just gets transversed to the other ball the first ball comes to rest
i.e. V1f=0
V2f=v
2)simple relation== (V2f-V1f)=(V2i-V1i)
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