1) A pair of bumper cars in an amusement park ride collide elastically as one ap

Question

1) A pair of bumper cars in an amusement park ride collide elastically as one approaches the other directly from the rear, as seen in part (a) of the figure below.

One has a mass of m1 = 446 kg and the other m2 = 556 kg, owing to differences in passenger mass. If the lighter one approaches at v1 = 4.46 m/s and the other is moving at v2 = 3.79 m/s,

a)calculate the velocity of the lighter car after the collision.

b)Calculate the velocity of the heavier car after the collision.

c)Calculate the change in momentum of the lighter car.

d) Calculate the change in momentum of the heavier car.

2)An 27.2 g rifle bullet traveling 211 m/s buries itself in a 3.44 kg pendulum hanging on a 2.86 m long string, which makes the pendulum swing upward in an arc. Determine the vertical and horizontal components of the pendulum's displacement. Give vertical component first.

3) You are witness to a traffic accident on the road in front of AUA.

An SUV with a mass of 2490 kg has stopped to wait for students crossing the street.
The driver of a 830 kg car is not paying attention and rear-ends the SUV.
As a result of the collision, both vehicles are stuck together and skid forward a distance of 1.3 meters.

Assume that after the collision, the tires of both vehicles are locked and that the coefficient of kinetic friction between the tires and the road is 0.85

Based on this information, calculate how fast the car was moving before it hit the SUV.

Explanation / Answer

Question 1 ....

1)
here
intial angular velocity (u) = 6540 rpm = 6540 * 2*pi / 60 rad / sec = 684.8672 rad / sec
time ( t) = 3.66 sec
final angular velocity (v) = 0

so angular acceleration = (v - u ) / t = ( 0 - 684.8672 )/3.66 = -187.122 rad / sec2

2)
angular velocity (u) = 2024 rpm = 2024 * 2*pi / 60 rad / sec = 211.9528 rad / sec

linear speed of the particle at the edge = angular velocity ( in rad / sec ) * distance from center = ( 211.9528) * ( 0.5 / 2 ) = 52.9882 m/sec

3)
Let us first calculate the angular velocity in rad / sec .. Let the angular velocty be w rad / sec
distance r = 7.01 cm = 0.0701 m
accelarion = 84000 g = 84000 * 9.8 = 823200 m/sec

accelaration = angular velocity^2 * distance
823200 = w^2 * 0.0701
so angular velocity w = 3426.8388 rad / sec
so angualr velocity in RPM = 3426.8388 * 60 / (2*pi ) = 32723.9 RPM

4)
intial angular velocity (u) = 4580 RPM = 4580 * 2*pi / 60 rad / sec = 479.6165 rad / sec
final angualr velocity ( v ) = 1140 RPM = 1140 * 2*pi / 60 = 119.3805 rad /sec
time (t) = 2.55 sec

so angular accleraion (a) = (v-u) / t = (119.3805 - 479.6165) / 2.55 = -141.26902 rad / sec2

Let the angle rotated be s ..
so.. s = u*t + 0.5*a*t^2 = (479.6165*2.55) + (0.5*(-141.26902) * 2.55^2) = 763.7212 radians ...

so no. of revolution = angle / (2*pi ) = 763.7212 / ( 2*pi ) = 60.775 revoluitons

5)
intial angular velocity (u) = 246 RPM = 246 * 2*pi / 60 rad / sec = 25.76106 rad / sec
final angualr velocity ( v ) = 372 RPM = 372 * 2*pi / 60 = 38.95575 rad /sec
time (t) = 6.02 sec

so angular accleraion (a) = (v-u) / t = (38.95575 - 25.76106) / 6.02 = 2.191809 rad / sec2

Let the angle rotated be s ..
so.. s = u*t + 0.5*a*t^2 = (25.76106*6.02) + (0.5*2.191809 * 6.02^2) = 194.7976 radians ...

so distance travelled by edge = angle * distance from center = 194.7976*0.307/2 = 29.9014 metres






Question 2

3) Intial angular speed (u) = 0 rad /sec
Final angular speed (v) = 14.2 RPM = 14.2 * 2*pi /60 rad / sec = 1.487 rad / sec
time (t) = 11.8 sec

so angular acceleration(a) = ( v- u ) / t = ( 1.487 - 0 ) / 11.8 = 0.1260 rad /sec^2

Inertia of the system = Inertia of the disc + inertia of two childre
I = (740 * 2.44^2 / 2) + 2*(23.8 * 2.44^2) = 2486.2234 kg-m^2

so troque = I * a = 2486.2234 * 0.1260 = 313.26415 N-m

so force required = troque / radius = 313.26415 / 2.44 = 128.387 N

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