Can you show me step by step how you reached the answer for these questions than

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Can you show me step by step how you reached the answer for these questions

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An empty freight car of mass 10,000 kg rolls at 4 m/s along a level track and collides with a loaded car mass 20,000 kg, standing at rest with brakes released. Friction can be ignored. If the cars stick together a) Find their speed after the collision. b) Find the change in kinetic energy of the two freight cars as a result of the collision. c) With what speed should the loaded car be rolling towards the empty car for both to be brought to rest by the collision? Two ice skaters, Daniel (mass 65.0 kg) and Rebecca (mass 45.0 kg) are practising. Daniel stops to tie his shoelace and, while at rest, is struck by Rebecca, who is moving at 14.0 m/s before she collides with him. After the collision, Rebecca has a velocity of magnitude 6.00 m/s at an angle of 40.1° from her initial direction. Both skaters move on a frictionless, horizontal surface of the rink. a) What are the magnitude and direction of Daniel s velocity after the collision? b) What is the change in the total kinetic energy of the two skaters as a result of the collision? (8 marks) A 0.7 kg object is dropped. Figure 3 shows its vertical velocity. Air resistance is a factor -5.0 -10.0 15.0 -20.0 25.0 30.0 -3S.0 2.0 4.0 6.0 t(s) Figure 3 a) Is the acceleration constant? Can you use equations for constant acceleration to model the object's motion? b) Using the figure, find an approximate value for the acceleration at t 4s c) Draw a free body diagram and determine the force of air resistance at t 4s (6 marks)

Explanation / Answer

1) a) m1*vi = (m1 + m2)*vf

vf = m1*vi/(m1 + m2) = 1.33 m/s

b) change in KE = 0.5(m1 + m2)*vf^2 - 0.5*m1vi^2 = 26533.5 J - 80000 J = -53466.5 J

c) m1*vi + m2*vf' = 0

vf' = -(m1/m2)*vi = 2 m/s

2)    Linear momentum is conserved.
Consider R's initial direction to be the x-axis.
vR1x = 14.0 m/s
vR2x = (6.00 m/s) cos(40.1 deg) = 4.59 m/s
vR2y = (6.00 m/s) sin(40.1 deg) = 3.86 m/s
Momentum components are
pR1x = (14.0 m/s)(45 kg) = 630 kg m/s
pR2x = (4.59 m/s)(45 kg) = 206.55 kg m/s
pR2y = (3.86 m/s)(45 kg) = 173.7 kg m/s

Thus, the components of Daniel's momentum after the collision are
pD2x = (630 - 206.55) kg m/s = 423.45 kg m/s
pD2y = -173.7 kg/ms
The components of Daniel's velocity after the collision are found by dividing out his 65-kg mass:
vD2x = 6.515 m/s
vD2y = -2.672 m/s

a) Magnitude of vD2 = sqrt(6.515^2 + 2.672^2) = 7.042 m/s
b) Direction of vD2 = arctan(-2.672/6.515)
= -22.3 degrees, i.e., 62.4 degrees from Rebecca's new direction, or 22.3 deg from her old direction (but turned in the opposite sense from the way she turned)

c) Kinetic energy: the 90-degree separation suggests that almost no kinetic energy was lost, but you should figure out (1/2)mv^2 for each skater after the collision, and subtract their sum from the original KE of (1/2)(45 kg)(14.0)^2 = 4410 J.

KE(D) = 0.5*65*7.042^2 = 1611.7 J

KE(R) = 0.5*45*6^2 = 810 J

Change = 4410 - 1611.7 - 810 = 1988.3 J

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