Please someone give me a hand with these as soon as possible? These are due at 1
ID: 1418921 • Letter: P
Question
Please someone give me a hand with these as soon as possible? These are due at 11:55 tonight and the questions I've found similar aren't helping me.
1. The drawing shows a version of the loop-the-loop trick for a small car. If the car is given an initial speed of v = 4.1 m/s, what is the largest value that the radius r can have if the car is to remain in contact with the circular track at all times? (I can't paste the picture here)
2. A car of mass 1510 kg traveling at 26.0 m/s is at the foot of a hill that rises 115 m after travelling 4.4 km. At the top of the hill, the speed of the car is 6.0 m/s. Find the average power delivered by the car's engine, neglecting any frictional losses.
3. A cue ball traveling at 0.70 m/s hits the stationary 8-ball, which moves off with a speed of 0.23 m/s at an angle of 40° relative to the cue ball's initial direction. Assuming that the balls have equal masses and the collision is inelastic, what will be the speed of the cue ball?
Please help me! SOS!
Explanation / Answer
1) The initial KE must be equal to the KE at the top + the GPE gained in climbing to the top of the loop.
And at the top the velocity must be g.R
( because the centripetal force (mV^2/R) must be completely supplied by the gravity force (mg) at this point .)
So
mV^2 / 2 = m (gR )^2 / 2 + 2 m g R
8.405 = gR / 2 + 2 g R
5gR/2 = 8.405
R = 0.343 m
2) PE at start of climb = 1/2 (mv^2) = 1/2*1510*26^2 = 510380 J
E at top of hill = 1/2 (mv^2) + (mgh) =27180 +1701770 = 1728950 J
Difference =1218570 J provided by engine.
The car's average speed is = 26+6/2 = 32/2 = 16 m/sec
distance = 4400 m
time to travel = 4400/16 = 275 sec
power = energy/time = 1218570/275 = 4431.16 watt
3) Use the law of conservation of momentum
M1V1 + M2V2 = M1V3 + M2V4
where
M1 = mass of the cue ball
V1 = initial velocity of the cue ball = 0.70 m/sec
M2 = mass of the 8-ball
V2 = initial velocity of the 8-ball = 0 (ball is at rest)
V3 = final velocity of cue ball
V4 = final velocity of 8-ball
Since M1 = M2, the above momentum equation simplifies to
V1 = V3 + V4
The X-component of the velocities will be
V1 = V3(cos A) + V4(cos 40)
where
A = angle at which the cue ball is deflected
Substituting values,
0.7 = V3(cos A) + 0.23(cos 40)
V3(cos A) = 0.524 (1)
The y-component of the velocities will be
V3(sin A) = V4(sin 40)
V3(sin A) = 0.23(sin 40) = 0.148 (2)
Divide Equation 2 by Equation 1,
tanA = 0.282
A = 15.77 degrees
and using Equation 1,
V3(cos 15.77) = 0.524
V3 = 0.545 m/sec