1 An object is thrown upward from the edge of a tall building with a velocity of 10 m/s. Where will the object be 3 s after it is thrown? Take
g=10ms−2
15 m above the top of the building
30 m below the top of the building
15 m below the top of the building
30 m above the building
2 A stone thrown from ground level returns to the same level 4 s after. With what speed was the stone thrown? Take
g=10ms−2
20 m/s
10 m/s
30 m/s
15 m/s
3 What is common to the variation in the range and the height of a projectile?
horizontal velocity
time of flight vertical velocity, horizontal acceleration
vertical velocity
horizontal acceleration
4 A cart is moving horizontally along a straight line with constant speed of 30 m/s. A projectile is fired from the moving cart in such a way that it will return to the cart after the cart has moved 80 m. At what speed (relative to the cart) and at what angle (to the horizontal) must the projectile be fired?
35.8 m/s at 24 degrees
38.6 m/s at 54 degrees
27 m/s at 35 degrees
24 m/s at 44 degrees
5 The trajectory of a projectile is
an ellipse
a circle
a parabola
a straight line
6 The motion of a ball rolling down a ramp is one with
constant speed
increasing acceleration
constant acceleration
decreasing acceleration
7 How fast must a ball be rolled along the surface of a 70-cm high table so that when it rolls off the edge it will strike the floor at the same distance (70cm) from the point directly below the edge of the table?
174.5 cm/s
185.2 cm/s
215.3 cm/s
143.7 cm/s
8 A ball is kicked and flies from point P to Q following a parabolic path in which the highest point reached is T. The acceleration of the ball is
zero at T
greatest at P
greatest at T and Q
the same at P as at Q and T
9 A mass accelerates uniformly when the resultant force acting on it
is zero
is constant but not zero
increases uniformly with respect to time
is proportional to the displacement of the mass from a fixed point
10 The term that best describes the need to hold the butt of a riffle firmly against the shoulder when firing to minimise impact on the shoulder is
forward displacement
forward acceleration
recoil velocity
pressure
11 Two trolleys X and Y with momenta 20 Ns and 12 Ns respectively travel along a straight line in opposite directions before collision. After collision the directions of motion of both trolleys are reversed and the magnitude of the momentum of X is 2 Ns. What is the magnitude of the corresponding momentum of Y?
6 Ns
8 Ns
10 Ns
30 Ns
12 A force of
2i⃗ +7j⃗
N acts on a body of mass 5kg for 10 seconds. The body was initially moving with constant velocity of
i⃗ −2j⃗ m/s
. Find the final velocity of the body in m/s, in vector form.
5i⃗ +12j⃗
12i⃗ −5j⃗
10i⃗ −7j⃗
7i⃗ +10j⃗
13 The exhaust gas of a rocket is expelled at the rate of 1300 kg/s, at the velocity of 50 000 m/s. Find the thrust on the rocket in newtons
6.5×107
3.5×107
7.6×107
5.7×107
14 Sand drops at the rate of 2000 kg/min. from the bottom of a hopper onto a belt conveyor moving horizontally at 250 m/min. Determine the force needed to drive the conveyor, neglecting friction.
500 N
800 N
139 N
152 N
15 A 30,000-kg truck travelling at 10.0m/s collides with a 1700-kg car travelling at 25m/s in the opposite direction. If they stick together after the collision, how fast and in what direction will they be moving?
8.1 m/s in the direction of the truck's motion
12.3 m/s in the direction of the car's motion
24.2 m/s in the direction of the car's motion
17.6 m/s in the direction of the truck's motion
16 A gun of mass M is used to fire a bullet of mass m. The exit velocity of the bullet is v. Find the recoil velocity of the gun
Mv/m
mv/M
–Mv/m
−mv/M
17 A 40-g ball travelling to the right at 30 cm/s collides head on with an 80-g ball that is at rest. If the collision is perfectly elastic, find the velocity of each ball after collision
the first ball is going to the right at 10m/s while the other is going to the left at 20m/s
the first ball is going to the left at 10m/s while the other is going to the right at 20m/s
the first ball is going to the left at 20 m/s while the other is going to the right at 10 m/s
the first ball is going to the right at 10 m/s while the other is going to the left at 10 m/s
18 A 10-g bullet of unknown speed is shot horizontally into a 2-kg block of wood suspended from the ceiling by a cord. The bullet hits the block and becomes lodged in it. After the collision, the block and the bullet swing to a height 30cm above the original position. What was the speed of the bullet? (This device is called the ballistic pendulum). Take
g=9.8ms−2
487 m/s
640 m/s
354 m/s
700 m/s
19 How large an average force is required to stop a 1400-kg car in 5.0 s if the car’s initial speed is 25 m/s?
2000 N
3500 N
9000 N
7 000N
20 Which of these is NOT a statement of Newton’s law of universal gravitation?
gravitational force between two particles is attractive as well as repulsive
gravitational force acts along the line joining the two particles
gravitational force is directly proportional to the product of the masses of the particles
gravitational force is inversely proportional to the square of the distance of the particles apart
1. V = Vo + g*t = 0 @ max ht.
10 - 10t = 0
Tr = 1 s. = Rise time or time to reach max. Ht.
Tf = Tr = 1 s. = Time to fall back to
top of bldg.
d = Vo*t + 0.5g*t^2
d = 10*(3-2) + 5*(3-2)^2 = 15 m. Below
top of blgd.
2. Tr + Tf = 4 s.
Tr = Tf = 2 s.
V = Vo + g*Tr = 0 @ max ht.
Vo - 10*2 = 0
Vo = 20 m/s.
3. Vertical velocity.
4. Dx = Xo * T = 80 m.
30 * T = 80
T = 2.666 s. = Time in flight.
T = Tr + Tf = 2.666 s.
Tr = Tf = 2.666/2 = 1.333 s.
Y = Yo + g*Tr = 0 @ max ht.
Yo = - g*Tr = 10*1.333 = 13.33 m/s = Ver
component of initial velocity.
Vo = Xo + iYo = 30 + i13.33 = 32.8 m/s
[24o].
5. A parabola.
8. Zero at T.
9. Is constant but not zero.
15. (30,000*10) - (1700*25) = 257,500
m*V = 257,500
(30000+1700)V = 257,500
V = 8.1 m/s.
19. V = Vo + a*t = 0
25 + a*5 = 0
5a = -25
a = -5 m/s^2.
F = m*a = 1400 * -5 = -7000 N. = 7000 N. in opposite direction of the motion.
1. To solve this problem, you can use the equation for the vertical displacement of an object in free fall:
𝑦 = 𝑣_0𝑡 + 1/2𝑔𝑡^2
where 𝑦 is the vertical displacement, 𝑣_0 is the initial velocity, 𝑔 is the acceleration due to gravity, and 𝑡 is the time.
In this case, the object is thrown upward, so the initial velocity is positive (+10 m/s) and the acceleration due to gravity is negative (-10 m/s^2).
Plugging in the values, we get:
𝑦 = (+10 m/s)(3 s) + 1/2(-10 m/s^2)(3 s)^2
= 30 m - 45 m
= -15 m
Since the displacement is negative, the object will be 15 m below the top of the building after 3 s. Therefore, the correct answer is 15 m below the top of the building.
2. To solve this problem, you can use the equation for the vertical displacement of an object in free fall:
𝑦 = 𝑣_0𝑡 + 1/2𝑔𝑡^2
In this case, the object is thrown from ground level, so the initial velocity is positive (+𝑣_0), and the acceleration due to gravity is negative (-10 m/s^2).
The object returns to the same level after 4 s, so the final displacement 𝑦 will be zero.
Plugging in these values, we have:
0 = 𝑣_0(4 s) + 1/2(-10 m/s^2)(4 s)^2
0 = 4𝑣_0 - 80
Solving for 𝑣_0:
4𝑣_0 = 80
𝑣_0 = 20 m/s
Therefore, the stone was thrown with a speed of 20 m/s.
3. The variation in the range and the height of a projectile is influenced by the projectile's time of flight, vertical velocity, horizontal acceleration, and horizontal velocity.
4. To solve this problem, you can use the equations of projectile motion to find the speed and angle at which the projectile should be fired.
First, find the time it takes for the projectile to return to the cart. Since the cart has moved 80 m, the time it takes for the projectile to cover this distance horizontally is t = 80 m / (30 m/s) = 2.67 s.
Next, use the equations for the horizontal and vertical components of velocity to find the initial speed and angle:
Horizontal component: 𝑣_𝑥 = 𝑣_0 cos(𝜃)
Vertical component: 𝑣_𝑦 = 𝑣_0 sin(𝜃)
The horizontal component of velocity remains constant at 30 m/s, so:
30 m/s = 𝑣_0 cos(𝜃)
The vertical component of velocity should be zero at the highest point of the projectile's trajectory:
0 = 𝑣_0 sin(𝜃) - 𝑔𝑡
0 = 𝑣_0 sin(𝜃) - 10 m/s^2 * 2.67 s
0 = 𝑣_0 sin(𝜃) - 26.7 m/s
Now, solve these two equations simultaneously to find 𝑣_0 and 𝜃:
From the first equation: 𝑣_0 = 30 m/s / cos(𝜃)
Substitute 𝑣_0 in the second equation: 0 = (30 m/s / cos(𝜃)) * sin(𝜃) - 26.7 m/s
0 = 30 tan(𝜃) -26.7
Solving for 𝜃 gives 𝜃 ≈ 24 degrees.
Now substitute this value of 𝜃 back into the first equation:
𝑣_0 = 30 m/s / cos(24 degrees) ≈ 35.8 m/s
Therefore, the projectile must be fired at a speed of approximately 35.8 m/s and at an angle of approximately 24 degrees to hit the cart when it returns after moving 80 m.
5. The trajectory of a projectile is a parabola.