Assignment 1
1. A student of mass 50 kg tests Newton's laws by standing on a bathroom scale in an elevator.
Assume that the scale reads in Newtons. Find the scale reading when the elevator is a)
accelerating upward at .5 m/s 2, b) going up at a constant speed of 3.0 m/s and c) going up
but decelerating at 1.0 m/s 2.
2.
A 10 kg box is attached to a 7 kg box which rests on a 30 o incline. The coefficient of kinetic
friction between each box and the surface is
system and b) the tension in the rope.
= .1 . Find a) the rate of acceleration of the
3.
A car travelling at a constant speed of 30 m/s passes a police car at rest. The policeman
starts to move at the moment the speeder passes his car and accelerates at a constant rate
of 3.0 m/ s 2 until he pulls even with the speeding car. Find a) the time required for the
policeman to catch the speeder and b) the distance travelled during the chase.
4.
A stone is thrown vertically upward from the edge of a building 19.6 m high with initial
velocity 14.7 m/s. The stone just misses the building on the way down. Find a) the time of
flight and b) the velocity of the stone just before it hits the ground.
5. How long would it take for a net upward force of 100. N, to increase the speed of a 50. kg
object from 100. m/s to 150. m/s.
6.
Initially a soccer ball is going 23.5 m/s, south. In the end, it is travelling at 3.8 m/s, south.
The ball's change in momentum is 17.24 kg m/s, north. Find the ball's mass.
7. A 62.0 kg curler travelling at + 1.72 m/s runs into a 78.1 kg curler and the 62.0 kg
curler comes to a stop. If the 78.1 kg curler was originally moving at + 0.85 m/s, find his
velocity after the interaction.
8.
A 62.0 kg curler runs into a stationary 78.1 kg curler and they hold on to each
other. Together they move away at 1.29 m/s, west. What was the original velocity of the
62.0 kg curler?
9. A 84.0 kg (total mass) astronaut in space fires a thruster that expels 35 g of hot gas at 875
m/s. What is the velocity of the astronaut after firing the shot?
10. Find the real number b so that vectors A and B given below are perpendicular
A = (-2 , -b) , B = (-8 , b).
NOTE
Scalar Product of Vectors
The scalar product (also called the dot product and inner product) of vectors A and B is
written and defined as follows
Fig1. - Angle between vectors and scalar product.
A·B = | A | | B | cos (θ)
where | A | is the magnitude of vector A , | B | is the magnitude of vector B and θ is the angle
made by the two vectors. The result of a scalar product of two vectors is a scalar quantity.
For vectors given by their components: A = (Ax , Ay, Az) andB = (Bx , By, Bz), the scalar
product is given by
A·B = AxBx + AyBy + AzBz
Note that if θ = 90°, then cos(θ) = 0 and therefore we can state that: Two vectors, with
magnitudes not equal to zero, are perpendicular if and only if their scalar product is equal to
zero.
The scalar product may also be used to find the cosine and therefore the angle between two
vectors
cos (θ) = A·B / | A | | B |
Properties of the Scalar Product
1) A·B = B·A
2) A· (B + C) = A·B + A·C
11. Given vector U = (3 , -7), find the equation of the line through point B(2 , 1) and
perpendicular to vector U.
12. A point M(x , y) is on the line through point B(2 , 1) and perpendicular to vector U = (3 , -7)
if and only if the vectors BM and U are perpendicular.
13. An object is launched at a velocity of 20 m/s in a direction making an angle of 25°
upward with the horizontal.
a) What is the maximum height reached by the object?
b) What is the total flight time (between launch and touching the ground) of the
object?
c) What is the horizontal range (maximum x above ground) of the object?
d) What is the magnitude of the velocity of the object just before it hits the ground?
SOLUTION TO Qn.13 SAVES AS AN EXAMPLE
a) The formulas for the components Vx and Vy of the velocity and components x and
y of the displacement are given by
Vx = V0 cos(θ)
Vy = V0 sin(θ) - g t
x = V0 cos(θ) t
y = V0 sin(θ) t - (1/2) g t2
In the problem V0 = 20 m/s, θ = 25° and g = 9.8 m/s2.
The height of the projectile is given by the component y, and it reaches its maximum
value when the component Vy is equal to zero. That is when the projectile changes
from moving upward to moving downward.(see figure above) and also the animation
of the projectile.
Vy = V0 sin(θ) - g t = 0
solve for t
t = V0 sin(θ) / g = 20 sin(25°) / 9.8 = 0.86 seconds
Find the maximum height by substituting t by 0.86 seconds in the formula for y
maximum height y (0.86) = 20 sin(25°)(0.86) - (1/2) (9.8) (0.86) 2 = 3.64 meters
b) The time of flight is the interval of time between when projectile is launched: t1
and when the projectile touches the ground: t2. At t = t1 and t = t2, y = 0 (ground).
Hence
V0 sin(θ) t - (1/2) g t2 = 0
Solve for t , t(V0 sin(θ) - (1/2) g t) = 0
two solutions t = t1 = 0 and t = t2 = 2 V0 sin(θ) / g
Time of flight = t2 - t1 = 2 (20) sin(θ) / g = 1.72 seconds.
c) In part c) above we found the time of flight t2 = 2 V 0 sin(θ) / g. The horizontal
range is the horizontal distance given by x at t = t2.
range = x(t2) = V0 cos(θ) t2 = 2 V0 cos(θ) V0 sin(θ) / g = V02 sin(2θ) / g = 202 sin
(2(25°)) / 9.8 = 31.26 meters
d) The object hits the ground at t = t2 = 2 V0 sin(θ) / g (found in part b above)
The components of the velocity at t are given by
Vx = V0 cos(θ)
Vy = V0 sin(θ) - g t
The components of the velocity at t = 2 V0 sin(θ) / g are given by
Vx = V0 cos(θ) = 20 cos(25°)
Vy = V0 sin(25°) - g (2 V0 sin(25°) / g) = - V0 sin(25°)
The magnitude V of the velocity is given by
V = √[ Vx2 + Vy2 ] = √[ (20 cos(25°))2 + (- V0 sin(25°))2 ] = V0 = 20 m/s
14. A ball is kicked at an angle of 35° with the ground.
a) What should be the initial velocity of the ball so that it hits a target that is 30
meters away at a height of 1.8 meters?
b) What is the time for the ball to reach the target?
1. A projectile is to be launched at an angle of 30° so that it falls beyond the pond of
length 20 meters as shown in the figure.
a) What is the range of values of the initial velocity so that the projectile falls between
points M and N?