Free Force and Laws of Motion 01 Practice Test - 9th Grade
Question 1
A car of mass 1,000 kg is moving at a speed of 20 ms−1.
It is brought to rest at a distance of 50 m. Calculate the net force acting on the car.
4,000 N
3,000 N
2,000 N
1,000 N
SOLUTION
Solution : A
For calculating force, we need to calculate deceleration first. From the third equation of motion,
v2−u2=2aS where, v- final velocity, u- initial velocity, a- acceleration and S- distance travelled.
Given, u=20 ms−1, S=50 m, mass m=1000 kg and v=0.
Therefore, 0−202=2×a×50
⇒a=−4 ms−2
Let the force acting on the car be F.
From Newton's second law,
F=ma
⇒F=1000×(−4)=−4000 N
Hence, the net force is −4,000 N. The negative sign indicates that the force is acting opposite to the direction of initial velocity and causes deceleration. This net force is actually the frictional force acting between the ground and the tyre of the car.
Question 2
If A and B are two objects of masses 6 kg and 3 kg respectively, then the inertia of object A is more than that of object B.
SOLUTION
Solution : A
Inertia is the natural tendency of an object to resist a change in its state of motion or of rest. Quantitatively, the inertia of an object is measured by its mass. More the mass more will be its inertia. As object A is heavier than object B, it has more inertia.
Question 3
When a bullet is fired from a pistol, in which direction the pistol moves (ignore the force exerted by hand)?
SOLUTION
Solution : A
The total momentum of the bullet-pistol system was initially zero as both of them were at rest with respect to ground. During the shooting of the bullet, net external force on the gun and bullet system is zero. Hence, according to the law of conservation of momentum, the total momentum of the system will be zero even after firing the shot. Since the momentum of the bullet will be in forward direction, in order to balance it, the momentum of the pistol should be in backward direction or the pistol should move backwards.
Question 4
When a football player kicks a ball, he experiences a force on his leg which is equal to the force he exerted on the ball.
True
False
SOLUTION
Solution : A
According to Newton's third law, when a body A exerts a force on another body B, B also exerts an equal and opposite force on A. Hence, while kicking a ball, the footballer's leg will experience an equal amount of force but in opposite direction to that of the force he exerted on the ball.
Question 5
Calculate the force required to accelerate a car from rest to a velocity of 30 ms−1 in 10 seconds. The mass of the car is 1500 kg.
SOLUTION
Solution : C
Here, mass m = 1500 kg
Initial velocity, u = 0, final velocity, v = 30 ms−1, time taken, t = 10 s.
Now, putting these values in the first equation of motion,v=u+at, we get
30=0+a×t,
⇒ 10a=30
⇒ a=3 ms−2
Now, putting m= 1500 kg and a = 3 ms−2 in equation, F=m×a, we get
F=1500×3 = 4500 N.
Thus, the force required is 4500 N.
Question 6
A water tanker filled up to two-thirds of its height is moving with a uniform speed. On sudden application of brake, the water in the tank will _____.
SOLUTION
Solution : B
When the tanker is in motion, the water inside will also be in motion. When the tanker comes to an abrupt stop, the water's inertia of motion resists this sudden change in motion. As a result, the water will move forward, before eventually coming to rest.
Question 7
Jaikishan and Arun are pushing each other as shown in the given figure. Jaikishan’s push is weaker than that of Arun. As a result, they move in the direction shown in the figure.
Jaikishan and Arun move in the given direction because of:
Tension force
Unbalanced force
Gravitational force
Balanced force
SOLUTION
Solution : B
It is clear from the given information that Arun is pushing harder, i.e., their pushes are unequal. Hence, the net force between them will be unbalanced and the direction of force will be as shown in the given figure.
Question 8
A cannon ball of mass 2 kg is fired at a target 500 m away. The 400 kg cannon recoils with a velocity of 2 ms−1. Assuming cannon ball moves with constant velocity, when will it hit the target?
1.25 s
2 s
1.5 s
0.2 s
SOLUTION
Solution : A
Let mass of cannon be m1 = 400 kg
Let mass of cannon ball be m2 = 2 kg
Let initial velocities of the cannon and cannon ball be u1 and u2 respectively.
Let final velocity of cannon and cannon ball be v1 and v2 respectively.
From the law of conservation of momentum,
total momentum before firing = total momentum after firing.
∴ m1u1+m2u2=m1v1+m2v2
As the cannon and cannon ball is at rest initially, the initial velocity will be zero, i.e
0=m1v1+m2v2
Substituting the values:
0=400×(−2)+2×v
⇒v=400×22= 400 ms−1
Hence, the velocity of the cannonball will be 400 ms−1
Since we know,
time=displacementvelocity
Thus, time taken to hit the target= 500400 = 1.25 s
Question 9
Which of the following has SI unit as kg m s−2?
SOLUTION
Solution : A
From, Newton's second law of motion, we know , F=m×a where F is the force applied, m is the mass and a is the acceleration. The SI unit of mass is kg and acceleration is m s−2. So, the SI unit of force is kg m s−2, also known as Newton(N).
Question 10
An object moves with an acceleration of 3 m s−2 when a force of 30 N is applied. The mass of the body is:
SOLUTION
Solution : D
Given, force, F=30 N and acceleration, a=3 m s−2.
Let m be the mass of the body.
We know, F=ma (Newtons's second law of motion)
∴ m=Fa=303=10 kg
Therefore, mass of the body is 10 kg.