I need help on these phyiscs problems asap Thank you!! :) 1. A car of mass 1560

Question

I need help on these phyiscs problems asap Thank you!! :)

1. A car of mass 1560 kg traveling at 24 m/s is at the foot of a hill that rises 105 m in 3.8 km. At the top of the hill, the speed of the car is 8 m/s. Find the average power delivered by the car's engine, neglecting any frictional losses -----Watts

2. 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 = 4.6 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?

3.The drawing shows a collision between two pucks on an air-hockey table. Puck A has a mass of 0.036 kg and is moving along the x axis with a velocity of 5.5 m/s. It makes a collision with puck B, which has a mass of 0.061 kg and is initially at rest. The collision is not head-on. After the collision, the two pucks fly apart with the angles shown in the drawing. Find the final speed of puck B.

4. The drawing shows a collision between two small balls. Ball 1 has a mass of 140 g and is moving along the x axis with a velocity of 5.40 m/s. It makes a collision with ball 2, which has a mass of 210 g and is initially at rest. After the collision, the two balls fly apart with angles as shown in the figure below (? = 60? and ? = 58?). What is the magnitude of the velocity of ball 2 after the collision?

5. A cue ball traveling at 0.75 m/s hits a stationary 8-ball, which moves off with a speed of 0.25 m/s at an angle of 30? relative to the cue ball's initial direction. Assuming that the balls have equal masses and the collision is inelastic, at what angle will the cue ball be deflected

Explanation / Answer

1. without frictional losses, we can write

KE at bottom = PE at top +KE at top + work done by engine

1/2 *1560 kg *(24m/s)^2 = 1560 kg x 9.8m/s/s x 105 m +1/2 * 1560 kg *(8m/s)^2 + work

hence work = 1.205880x10^6J

power = work/time so we need to find the time it took to get up the hill

the average speed = 16m/s, so it takes a time of 3800m/16m/s = 237.5 s to get up the hill, this means the power supplied was

power = 1.205880 x10^6J/237.5 s=5077.38 W Ans.

2. At the top of the loop, there is a gravitational force pulling down on the car, and a normal force up. The car just stays on as the normal force approaches 0.
Fy = Fy
ma[c] = Fn - Fg
m v^2 / r = Fn - mg
m v^2 / r = mg
v^2 = rg
v = sqrt (9.8r)

E(initial) = E(final)
KE(initial) = PE(final) + KE(final)
1/2 mv^2 = mgh + 1/2 mv^2
1/2 (4.6m/s)^2 = (9.8)(2r) + 1/2(9.8r)
10.58 = 19.6r + 4.9r
10.58 = 24.5r
r = 0.4318 m Ans.

3. Use conservation of momentum

momentum in y initial = momentum in y final, because there is no initial momentum you can say that the sum of puck A's vertical momentum and puck B's vertical momentum are zero

so (.036)(v1)sin(65) + (.061)(v2)sin(323) = 0

set up the conservation of momentum in the x direction

(.036)(5.5) = (.036)(v1)cos(65) + (.061)(v2)cos(323)

solve for one of the variables, I solved for (v2)

((.036)(v1)sin(65))/((-.061)(sin323)) = v2

plug it into the second equation so

simplified it looks like this (.036)(5.5) =v1((.036)(cos65)-(cos(323)(.026)(sin(65...

plug that value back into one of the equations and solve for v2 to get

we can calculate the values of v1 & v2.

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