# Parabolic motion, work and kinetic energy, linear momentum, linear and angular motion – problems and solutions

1. A ball is thrown from the top of a building with an initial speed of 8 m/s at an angle of 20^{o} below the horizontal. The ball hits the ground 3 seconds later.

a) How far from the bottom of building the ball touches the ground

b) How high is the ball thrown from the place

c) How long the ball reaches a height of 10 m from the place of throwing

v_{ox} = v_{o} cos 20^{o }= 8 (0.939) = 7.512 m/s

v_{oy} = v_{o} sin 20^{o }= 8 (0.342) = 2.736 m/s

a) x

Horizontal movements are analyzed as uniform linear motion so that used the formula for uniform linear motion.

x = v_{ox} t

x = (7.512 m/s)(3 s)

x = 22.536 m

b) y

The movements of the vertical direction are analyzed as the free fall motion using the free fall motion formula.

Use this formula y = v_{o} t + ½ g t^{2}

In free fall motion, v_{o} = 0 so the formula becomes : y = ½ g t^{2}

y = ½ (9.8)(3^{2})

y = (4.9)(9)

y = 44 m

c) y = 10 m

Use the formula y = ½ g t^{2}

10 = ½ (9,8) t^{2}

10 = 4.9 t^{2}

t^{2 }= 10 / 4.9

t^{2 }= 2

t = 1.4 s

2. A dog train team pulls a mass of 100 kg load as far as 2 km above the horizontal surface at a constant speed. If the coefficient of friction between the train and snow is 0.15, determine:

a) the work done by the dog

Calculate the kinetic friction:

F_{k }= µ_{k} N = µ_{k} w = µ_{k} m g = (0.15)(100 kg)(9.8 m/s^{2}) =147 N

The speed of mass is constant so according to Newton’s law, the resultant force = 0. So F_{k }= F = 147 N

The work done by the dog:

W = F d = (147 N)(2000 m) = 294,000 J

b) Energy lost by friction

Energy lost by friction:

Work = energy, so:

W = F_{k} d = (147 N)(2000 m) = 294,000 J

3. The 150 N horizontal force is used to push the box 6 meters above the rough surface. If the box moves at a constant speed:

a) Work done by the force of 150 N

W = F d = (150 N) (6 m) = 900 J

b) Energy lost due to friction

The box moves at a constant speed so that the resultant force acting on the box = 0.

So the horizontal force = 150 N = kinetic friction.

Energy lost due to friction:

W = F_{k} s = (150 N)(6 m) = 900 J

c) The coefficient of kinetic friction

The formula of friction force:

F_{k }= µ_{k} N = µ_{k} w = µ_{k} m g

150 = µ_{k} m (10)

15 = µ_{k} m

µ_{k} = 15/m

m = mass of the box

µ_{k} = coefficient of kinetic friction

Kinetic friction works when objects move, static friction works when objects are still stationary.

4. A particle has a speed (3i-4j) m/s. Determine the momentum component on the x and y-axis and the magnitude of momentum.

Speed on the x axis: 3 m/s

Speed on the y axis: 4 m/s

Momentum on the x axis: m v = 3m

momentum on the y axis: m v = 4m

The magnitude of momentum:

p^{2} = (3m)^{2 }+ (4m)^{2}

p^{2} = 9m^{2 }+ 16m^{2}

p^{2} = 25m^{2 }

p = 5m

p is a symbol of linear momentum.

5. A car is accelerated homogeneous from stationary to 22 m/s for 9 seconds. If the tire diameter is 58 cm, specify:

a) the number of rounds of tires during the movement (assume no slip)

Calculate car acceleration:

v_{t }= v_{o} + a t

22 = 0 + a 9

22 = 9a

a = 22/9 = 2.44

Calculate distance:

d = v_{o} t + ½ a t^{2} = 0 + ½ a t^{2}

d = ½ a t^{2 }= ½ (2.44)(9^{2}) = (1.22)(81) = 98.82 m

Distance of car = 98.82 meters = 9882 cm

Diameter of tire = 58 cm

The number of revolution = 9882 cm / 58 cm = 170.38 rev

b) The tire’s final rotation speed in rev/minute

Final speed = 22 meters/second = 2200 cm/second

1 rev = 1 diameter of tire = 58 cm

Number of revolution:

2200 cm / 58 cm = 40 revolutions

So the final speed in revolution/minute

40 rev/second = 40 rev / 60 seconds = 0.67 rev/minute or 0.67 rev/minute