Force and Motion Practice Test 4
Force and Motion Practice Test #4 Number of MCQS: 50 Difficulty Level Moderate 70 Percent and Difficult 30 Percent
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Category: Force & Motion Topical (FAM)
If the horizontal range of a projectile is maximum then the angle of the projectile must be ______ with horizontal.
2 / 50
When an object moves on a circular path and come back to its initial position, then:..
When a body reach to its initial Position, its Displacement become Zero, so Similarly incase of Circular Motion a body Must reach to its initial Position, so its Displacement Must be zero.
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Select One With maximum inertia?
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Newton’s second law of motion _______
Newton’s Second Law:
P=mv
Where F = Force, m = mass and a = acceleration
From above it is clear that Newton’s second law of motion explains about change in momentum.
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The acceleration of a moving body can be found from
Area under velocity time graph gives displacement and slope of the same graph gives acceleration
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If a car at rest accelerates uniformly to a speed of 144 km/h in 20 s, it covers a distance of
7 / 50
area under the force-time curve
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body, staring from rest and moving with constant acceleration, covers 10 m in the first second. Find the distance travelled by it in the second second
Using the equation for the distance travelled in the nth second,
we have, for the first second,
The distance travelled in the second second is
9 / 50
Which of the following should be constant for a body to have a constant momentum?
For a given mass P∝v If the momentum is constant then it's velocity must have constant, Acceleration and Force will remain zero.
10 / 50
The total quantity of motion contained in the same body is called _________.
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the distances travelled in successive equal time intervals will be in the ratio
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A billiard ball collides with a stationary identical billiard ball to make it move. If the collision is perfectly elastic, the first ball
Let m be the mass of the two identical balls.
From the conservation of momentum,
mu1 + mu2 = mv1 + mv2
⇒ mu1 = mv1 + mv2 ⇒ u1 = v1 + v2
In an elastic collision, the kinetic energy of the system before and after collision remains same.
After Simplification we left with:
2V1V2=0
13 / 50
As we move Forward during Swimming it is because of
It gives the basics idea of force.
Force (F) = Mass (m) × Acceleration (a)
It gives the formula and final definition of the force.
14 / 50
When a projectile is fired at an anlge Θ with respect to horizontal with velocity u, the horizontal component, ignoring air resistance
Because there is no acceleration or retardation along horizontal direction, hence horizontal component of velocity remains same
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When do we get maximum height in a simple projectile motion?
The formula for horizontal range is h = (v sinθ)2/2g. This will be maximum when sin θ = 1, which implies that θ = 90°. Hence the correct answer is when θ = 90°
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When a body is moving with a constant acceleration, the displacement-time graph is a
17 / 50
When a body A , is dropped freely from the top of the tower and another body B is projected horizontally from the same point
As the body A is dropped from rest,
As the body B is given a horizontal velocity at the time of release, it is going to follow the same trajectory as a body on a projectile motion, projected with a velocity having the same horizontal component as the horizontal velocity given to B at H. So the time taken by B from H to Y is same as that from X to H. Now let the vertical component of velocity at X be v then at H,
18 / 50
A cricket ball is thrown with a velocity of 15 m/s at an angle of 30o with the horizontal. The time of flight of the ball will be
Here g=10, u=15 m/s, θ with the horizontal is 30 °
19 / 50
A body starts to fall freely under gravity. The distances covered by it in first, second and third second are in ratio
According to equation of motion, distance covered in nth sec
a/2 is constant in all So it will Cancel out
20 / 50
Two bodies P and Q move in a straight line. Their position–time graphs are straight lines through the origin. The line for P is steeper than the line for Q. Which is correct?
On a position–time graph, the slope of the line gives speed.Steeper line means larger slope, so higher speed.Since P’s line is steeper than Q’s, P moves faster than Q.
21 / 50
Taking rocket + fuel as system total linear momentum must be conserved since there is no external force. As the exhaust moves out with high speed towards the Earth , the rocket moves with same momentum away from the earth as the net momentum should be zero .
22 / 50
When a body is accelerated
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car travels for a certain time. Its speed during the first half time is 20 m/sec and that during the second half time is 40 m/sec . Find the average speed.
Let the total time be 2t. Then the distance covered during the first half time t is S1=v1t, and that during the second half time is S2= v2t
24 / 50
The speeds of an object at the ends of 4 successive seconds are 20, 25, 30, and 35 meters per second, respectively. The acceleration of this object is
As we Know Chang in Velocity with respective time is Acceleration, Look at the Question the change is 5 and and time is 1 Sec: so 5 m/sec2
25 / 50
when we shake a carpet dust remove from it, which law of motion describe this
26 / 50
ball is projected with a speed of 10 m/s. What are the two angles of projection for which the range is 5.0 m ? (g = 10 m/s2)
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Football of mass 2 Kg was initially in hand of Ali Raza. He Throws the Ball with 20 m/s then find the impulse imparted to the ball
Given that: Mass of the ball (m) = 2 kg
Initial velocity of ball (V1) = 0 and Final velocity of ball (V2) = 20 m/s
Impulse (J) imparted to the ball = P2 – P1 = 2 × 20 – 2 × 0 = 40 kg m/s
28 / 50
Due to an acceleration of 2 m/s2, the velocity of a body increases from 20 m/s to 30 m/s in a certain period. Find the displacement (in m) of the body in that period.
Given - Acceleration (a) = 2 m/s2, Final velocity (v) = 30 m/s and Initial velocity (u) = 20 m/s
Displacement = s
We know,
⇒ v2 - u2 = 2as (Equation of Motion)
⇒ s = (v2- u2)/ (2 x a)
⇒ s = (302-202) / (2 x 2)
⇒ s = 500 /4
⇒ s = 125 m
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The angle to projection for which the maximum height and the horizontal range of a projectile are equal is
30 / 50
The SI unit of linear momentum is
as we know: p=mV=kg . m/sec
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True about Action and reaction
Newton's third law of motion : For every action there is always an equal and opposite reaction acting on the other body.
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A ball is projected up vertically with a velocity of 9.8 m/s. The time it takes to reach the ground is
Time to reach max height =u/g
Total time of flight =2u/g
Given:
u = 9.8 m/s
Time of flight = u/g=9.8/9.8=1 sec
Total time of flight=2 x 1 = 2 sec
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When particle moves the first half of a distance at a speed of 5 m/s and second half of the distance at speed 15 m/sec then average velocity is
When particle moves the first half of a distance at a speed of v1 and second half of the distance at speed v2
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If a body is thrown upwards with a velocity 20 m/sec, then neglecting the effect of air resistance, Total time taken to reach the top and then return to the ground
Total time taken to reach the top and then return to the ground: t = 2u/g
u=20 m/sec, and g= 10
t= 2(20)/10=4 Sec
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A tennis ball strikes a wall with a velocity v and rebounds back with the same velocity. What is the change in kinetic energy of the ball?
36 / 50
Newton’s first law of motion is not valid for______
37 / 50
A particle moving in a straight line covers half the distance with speed of 3 m/s. The other half of the distance is covered in two equal time interval with speed of 4 m/s and 5 m/s respectively. The average speed of the particle during this motion is
38 / 50
A cricket ball is hit at 30o with the horizontal with kinetic energy K. The kinetic energy at the highest point is
Given, θ=30°
KE=K. We know that,
As we Know that K:E at Highest Point during Projectile motion is:
39 / 50
Alina Through a ball Upward in vacuum at 100 m/s. how many time it take to reach the highest point is
By using the equation: v = u +at we can get the answer.
At highest point, velocity, v = 0 m/s and acceleration, a = -g = -10 m/s^2.
Initially the velocity, u = 100 m/s.
Therefore, 0 = 100 - 10t
=> t = (-100)/(-10) =10 seconds.
40 / 50
What is the approximate weight of a 7.5-kilogram mass sitting on the surface of Earth?
If we approximate the acceleration due to gravity as g = 10 m/s2, then the weight of a mass on the surface of Earth is given by Fg = mg = (7.5 kg) (10 m/s2) = 75 N.
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A ball of mass m1 and another ball of mass m2 are dropped from equal height. If the time taken by the balls are t1 and t2 respectively, then
in Free fall motion the time of flight is always independent of mass
42 / 50
When a projectile is fired at an angle Θ w.r.t. horizontal with velocity u, then its vertical component
The vertical component goes on decreasing and eventually becomes zero.
43 / 50
A straight horizontal line above the time axis on a velocity–time graph shows:
Horizontal v–t line means velocity value is same at all times.No change in velocity means zero acceleration.So the body moves with constant velocity.
Why others are wrong:B) Constant acceleration needs a sloping line, not flat.C) Decreasing velocity would be a line sloping downwards.D) At rest would be a horizontal line on the time axis at v = 0.
44 / 50
Displacement of sun with respect to earth is
As Earth is Revolving around the sun, The shortest distance between both at Each Point is "r"
The displacement of the Sun with respect to the Earth is the distance between the two objects. Since the Earth orbits the Sun at an average distance of r, the displacement is r. This is the typical way it is expressed, assuming a simplified model of the Earth's orbit as circular.
45 / 50
An object with a constant speed
a body can have a constant speed and still be accelerating. Acceleration can be either due to change in speed or due to direction of motion or both. Consider an example of a uniform circular motion. In a circular motion, the body moves with a constant speed but we still say that it is accelerating due to the change in direction
46 / 50
Force exerted on a body changes it's
A body acted upon by a certain force produces acceleration i.e. it undergoes change in it's velocity and momentum and kinetic energy of a moving body depends on the velocity of the body
47 / 50
If air resistance is not considered in projectiles, the horizontal motion takes place with
In the absence of air resistance, the projectile moves with constant horizontal velocity because acceleration due to gravity is totally vertical and its horizontal component is zero
48 / 50
If 5 bullets each of mass 3 Kg and velocity 2 m/s (per second) are fired from a gun then the average force acting on the gun is
If n bullets each of mass m and velocity v are fired from a gun then the average force acting on the gun is mnv
49 / 50
When a bus starts suddenly, the passengers are pushed back. This is an example of which of the following?
Laws of Motion given by Newton are as follows:
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On calculating which of the following quantities, the mass of the body has an effect in simple projectile motion?
All mentioned quantities except force are kinematic quantities. Force is a kinetic quantity. Force = m x a. Hence, the mass of the body has an effect on force calculation.
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