Physics Review Answer Key

Part C test review
1. A person walks 45. meters due south and then walks 30. meters due north. The entire trip takes the
person 10. minutes. Determine the magnitude and the direction of the person's total displacement.
Answer: 15 meter due south
2. The instant before a batter hits a 0.24-kilogram baseball, the velocity of the ball is 42 meters per second
west. The instant after the batter hits the ball, the ball's velocity is 35 meters per second east. The bat and
ball are in contact for 1.0 x 10-2 second.
(a) Determine the magnitude and direction of the average acceleration of the given baseball while it is in
contact with the bat.
(b) Calculate the magnitude of the average force the bat exerts on the given ball while they are in contact.
[Show all work,including the equation and substitution with units.]
(a) 7.7 x 10 3 m/s 2 east 42-(-35)/1x10 -2 ;
(b) 1.1 x 10 3 N
WORK SHOWN: a = , Fnet = ma, Fnet = (0.24 kg)(7.7 x 10 3 m/s2) = 1.8 x 10 3 N OR J = Fnet = hp,
Fnet = = 1.8 x 10 3 N
3. A 65-kilogram athlete jogs 2.8 kilometers along a straight road in 1.2 x 10 3 seconds.
(a) Determine the average speed of the athlete in meters per second.
(b) Calculate the average kinetic energy of the given athlete. [Show all work, including the equation and
substitution with units.]
(a) 2.3 m/s; 2800 m/1.2 x 10 3 seconds = 2.3 m/s
(b) 177 J
WORK SHOWN: KE = ½mv2 = ½ (65 kg)(2.3 m/s)2 = 177 J
4. On a snow-covered road, a car with a mass of 1.5 x 10 3 kilograms collides head-on with a van having a
mass of 2.3 x 10 3 kilograms traveling at 5.0 meters per second. As a result of the collision, the vehicles lock
together and immediately come to rest.
Calculate the speed of the car immediately before the collision. [Neglect friction.] [Show all work,
including the equation and substitution with units.]
7.7 m/s
WORK SHOWN: Pbefore = Pafter, m1v1 + m2v2 = 0, v1 = 2.3 x 10 3 5.0 , v1 = = 7.67 m/s OR 7.7 m/s
1.5 x 10 3
5. Force A with a magnitude of 6.5 newtons and force B with a magnitude of 8.3 newtons act concurrently
on point P If the forces are at an angle of 35 degrees, determine the magnitude of the resultant force of
vectors A and B.
6. A 1,000-kilogram car moving at 15 meters per second collides with a 2,100-kilogram car that is waiting
at rest at a traffic light. After the collision, the cars lock together and slide. Eventually, the combined cars
are brought to rest by a force of kinetic friction as the rubber tires slide across the dry, level, asphalt road
surface.

a) Calculate the speed of the locked-together cars immediately after the collision described in the reading
passage. [Show all work,including the equation and substitution with units.]
4.8 m/s
WORK SHOWN:
Pbefore = Pafter, (m1v1 + m2v2)before = (m1 + m2)vafter, vafter = (1000 x 15)+(2100 x 0),
(1000 + 2100)
vafter = 4.8 m/s
b) Calculate the magnitude of the frictional force that brings the locked-together cars described in the
reading passage to rest. [Show all work, including the equation and substitution with units.].
32) 2.0 x 10 4
WORK SHOWN: Ff = mFN and FN = mg, Ff = mmg, Ff = (0.67)(1,000 kg + 2,100 kg)(9.81 m/s2) = 2.0 x
10 4
7. A spring in a toy car is compressed a distance, x. When released, the spring returns to its original length,
transferring its energy to the car. Consequently, the car having mass m moves with speed v.
Derive the spring constant, k, of the car's spring in terms of m, x, and v. [Assume an ideal mechanical
system with no loss of energy.] [Show all work, including the equations used to derive the spring constant.]
PEs = KE,1/2 kx2 = ½ mv2
k = mv2/ x2
8. Calculate the magnitude of the impulse applied to a 1.5-kilogram cart to change its velocity from 0.75
meter per second east to 2.00 meters per second east. [Show all work, including the equation and
substitution with units.]
1.9 Ns
WORK SHOWN: J = p = mv, J = (1.5 kg)(1.25 m/s) = 1.9 Ns
9. A roller coaster car has a mass of 360. kilograms. Starting from rest, the car acquires 3.13 x 105 joules of
kinetic energy as it descends to the bottom of a hill in 4.3 seconds.
a) Calculate the height of the hill in the situation described. [Neglect friction.] [Show all work, including the
equation and substitution with units.]
88.7. m
WORK SHOWN: KE = PE = mgh, 3.13 x 10 5
(360) (9.8)
b) Calculate the speed of the roller coaster car at the bottom of the hill in the situation described. [Show all
work, including the equation and substitution with units.]
(a) 41.6 m/s
WORK SHOWN: KE = ½ mv2, v2 = 2(3.13 x 105) , v = 41.6 m/s;
360
c) Calculate the magnitude of the average acceleration of the roller coaster car as it descends to the bottom
of the hill. [Show all work, including the equation and substitution with units.]
9.7 m/s2
WORK SHOWN: a =v/t 41.6/4.3 , a = 9.7 m/s2