Download KEY to Newton`s third law practice

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Newton’s Third law Practice
Answer the following questions. Use the value for the acceleration due to gravity of 9.8 m/s2 when
needed. Use significant figures when finding resultant or net forces:
1. Imagine you try to start your car on a very cold morning, like last week, and find that you battery
is dead. You must push the car to move it and get it started (assuming it is a stick car). Why can’t
you move the car by remaining comfortable inside the car and pushing on the dashboard?
You receive a push by the car when you push the car. Both forces are equal in magnitude but act
in opposite directions. If you remain in the car while you push, you and the car form a system.
Since both forces act on the same system, the forces cancel out. When you push from the
outside, you and the car are independent from each other, so the equal in magnitude forces, but
opposite in direction do not cancel out.
2. We know the Earth pulls on the Moon (this gravitational force is the centripetal force that keeps
the moon on an elliptical trajectory). Does it follow that the Moon also pulls on Earth?
Yes, the Moon pulls on the Earth with the same force as the Earth pulls the Moon, but in opposite
direction. Both celestial objects rotate around their common center of mass. Because Earth is so
more massive than the Moon, the common center of mass lays somewhere on Earth.
3. A car accelerates along the road. Identify the force that changes the car’s velocity.
The road applies a force on the car that makes is accelerate. As the tires spin faster, they apply a
force on the road and therefore the road applies a force back on the car. Friction between the
tires and the road allow the forces to exist. Without friction, the car would not be able to apply
the force on the road, and the car would not be able to receive the force from the road.
4. A car and an innocent bug have a head collision. The force of the impact splatters the poor bug
over the windshield, and the car does not even notice. If the force applied by the car on the bug
is the same as the force applied by the bug on the car, how come the bug’s acceleration (change
in velocity) is much bigger than the car’s acceleration? (Think about Newton’s second law)
By Newton’s Second Law, we know = . We can solve for the acceleration so see that
= The bug’s acceleration (the change in the bug’s speed and/or direction) will be much
bigger than the car’s acceleration because, even though they receive a force of the same
magnitude, the mass of the bug is much smaller than the car’s mass. The car’s change in velocity
is imperceptible, but the bug’s change in velocity is very large.
5. In a rugby match between the New Zealand “All Blacks” and the Australia “wallabies” two
players pull the ball in opposite direction. The All Black player pulls with 200 N to the right, and
the Wallabies player with 150N to the left.
a. Sketch the free force diagram for the ball (forces acting on the ball)
-150
200 N
b. Find the resultant force, including its direction
Net force: 50 N to the rigth
6. A 0.110 kg ping pong paddle sits on top of the ping pong table at the end of the match. The
0.0027 kg ping pong ball sits motionless on top of the paddle. The table weighs around 1700. N.
a. Find the weights of the paddle and of the ball.
The weights of the paddle and the ball are the forces downward produce by gravity’s
acceleration on the mass of the paddle and the ball. We need Newton’s Second Law:
Given: mp=0.110kg, and mb=0.0027kg, where mp is the mass of the paddle and mb is the
mass of the ball. Negative values show a downward direction:
Paddle’s weight: Given = = 0.110 ∗ 9.8 = 1.078 = 1.1
Ball’s weigth: Given = = 0.0027 ∗ 9.8 = 0.02646 = 0.026
b. Calculate the magnitude and direction of the force exerted by the paddle on the ball.
Since the force exerted by the ball on the paddle is the ball’s weight (0.026N) pointing
down, the paddle exerts a force of the same magnitude but opposite direction on the
ball, so the force of the paddle on the ball is 0.026N (upwards).
c. Sketch the diagram of forces (free force diagram) acting on the paddle
Normal to the surface by the table on the paddle, which is the
sum of the ball and paddle’s weight: -0.026N + (-1.1 N) =-1.126 N
=-1.1N (sig figs)
0.026N N of ball’s weight
1.1N of paddle’s weight
d. Sketch the free force diagram for the ball,
Normal to the surface by the paddle, same magnitude as the ball’s weight
-0.026N N of ball’s weight
e. And sketch the free force diagram for the ping pong table.
Normal to the surface by the ground on the table: sum of the ball, the paddle’s and the
table’s weight: -0.026N + (-1.1 N) + (-1700N)=-1701.126 N =-1701.1N (sig figs), in opposite
direction = 1701.1N
-0.026N N of ball’s weight
Vectors are not to scale
-1.1N of paddle’s weight
-1700 N of table’s weight
7. A baseball is hit by a wimpy batter with a force that can be seen as being: 180.0 N at 40O from
the ground. The baseball has a mass of 0.145 kg. If the wind is strong enough to apply a force on
the ball of 2 N towards the South (or -2 N)
a. Determine the weight of the ball
ball’s weight: Given = = 0.145 ∗ −9.8
= −1.421 = −1.4 (2 sig figs)
With a frame of reference in which we chose to have negative numbers pointing down,
and positive pointing up; and positive to the North and negative to the South:
b. Sketch free force diagram for the ball
O
180.0 N at 40 by batter
40O
-2 N by wind
-1.4 N of ball’s weight
c. Find the net force in the horizontal (find the horizontal component of the force of the
bat on the ball )
Since cos40 =
F
40O
Fx
= ∗ !"#40 =180.0N*cos(40 = 137.89 = 137.9
The net force in the horizontal, while the bat is hitting the ball, is the sum of the force by
the wind (-2N) and the horizontal component of the batter’s force (137.9N):
-2N + 137.9 = 135.9N = 140N (or 140N to the North, rounded to no decimals)
d. Find the net force in the Vertical (find the vertical component of the force by the bat on
the ball)
Two forces act on the vertical, the weight of the ball and the vertical component of the
force applied by the batter on the ball. To find the vertical component of the batter’s
force:
Since sin40 =
F
40O
'
Fy
( = ∗ #)*40 =180.0N*sin(40 = 115.7
The net force in the vertical, while the bat is hitting the ball, is the sum of the balls’
weight (-1.4N) and the vertical component of the batter’s force (134.1N):
-1.4N + 115.7 = 114.3N (or 114.3N upwards)
e. Determine the force of the ball on the bat (include the direction with the sign)
The force of the ball on the bat has the same magnitude as the force of the bat on the
ball but opposite direction. It is 180.0N at 220O (40 O + 180 O).