Force and Laws of Motion CBSE Class 9 Physics Worksheets Set 07

Questions with Answers

Question 1. A body of mass 6 kg initially at rest is subjected to a steady force of 18 N. What is the acceleration caused in the body? In frictionless motion, what is the velocity acquired by the body in 10 seconds? Find its displacement in 10 s. What retardation is needed to stop the body within the next five seconds?
Answer:
Acceleration: \( a = \dfrac{F}{m} = \dfrac{18}{6} = 3 \) m/s².
Velocity after 10 s: \( v = at = 30 \) m/s. Displacement: \( s = \tfrac12 at^2 = 150 \) m.
Retardation to stop in 5 s: \( a = \dfrac{0 - 30}{5} = -6 \) m/s², that is 6 m/s².
The velocity-time graph is a straight line rising from 0 to 30 m/s in 10 s and a straight line falling back to zero in the next 5 s.
In simple words: Use F = ma, then the equations of motion for each stage.

Teacher's Note:
1. The velocity at the end of the first stage is the initial velocity of the second.
2. The area under the velocity-time graph (225 m in total) gives the total displacement.

Question 2. What is the weight in newtons exerted by a 10 kg mass on a horizontal stationary surface if g = 10 m/s²? How many newtons of force will it exert on a point of support that has a downward acceleration of 5 m/s²?
Answer:
On the stationary surface: \( W = mg = 10 \times 10 = 100 \) N.
On the accelerating support: \( W' = m(g - a) = 10(10 - 5) = 50 \) N.
100 N and 50 N.
In simple words: When the support falls with acceleration, the body presses less.

Teacher's Note:
1. The apparent weight decreases for a downward acceleration.
2. In free fall (a = g) the apparent weight is zero.

Question 3. A force of 100 N acts on 50 kg for 2 seconds. The same force acts on 25 kg for 2 seconds. Compare the momenta produced. Compare the accelerations caused.
Answer:
Momentum produced = impulse = F x t = 100 x 2 = 200 N s in both cases, so the ratio is 1 : 1.
Accelerations: \( a_1 = \dfrac{100}{50} = 2 \) m/s² and \( a_2 = \dfrac{100}{25} = 4 \) m/s², so the ratio is 1 : 2.
In simple words: The same impulse gives the same momentum; the lighter body accelerates twice as much.

Teacher's Note:
1. Impulse equals change of momentum.
2. Acceleration is inversely proportional to mass for the same force.

Question 4. A large force acting for a short time is called an impulsive force. The impulse produced by an impulsive force in 0.2 s is 1000 N s. Find the magnitude of the force.
Answer:
\( F = \dfrac{\text{impulse}}{t} = \dfrac{1000}{0.2} = 5000 \) N.
The force is 5000 N.
In simple words: Divide the impulse by the time for which the force acts.

Teacher's Note:
1. Impulse = F x t.
2. The shorter the time, the larger the force for a given impulse.

Question 5. A cannon ball (mass 2 kg) is fired at a target 500 m away. The 400 kg cannon recoils by 50 cm. What is the velocity of the cannon ball? Assuming that the ball moves with constant velocity, when will it hit the target? (The resistance offered by the ground to the cannon is 100 N.)
Answer:
Retardation of the cannon: \( a = \dfrac{100}{400} = 0.25 \) m/s². Recoil velocity: \( v_2^2 = 2as = 2 \times 0.25 \times 0.5 = 0.25 \), so \( v_2 = 0.5 \) m/s.
Conservation of momentum: \( 2v_1 = 400 \times 0.5 \Rightarrow v_1 = 100 \) m/s.
Time to hit the target = \( \dfrac{500}{100} = 5 \) s. Velocity = 100 m/s; time = 5 s.
In simple words: First find how fast the cannon moves back, then use momentum conservation.

Teacher's Note:
1. The momentum of the ball equals that of the cannon in magnitude.
2. Use distance = speed x time for the flight.

Question 6. A car moving with a velocity of 90 km/h takes 10 s to stop after the brake is applied. Calculate the force exerted by the brakes on the car if its mass with the driver is 500 kg.
Answer:
u = 90 km/h = 25 m/s, v = 0, t = 10 s, so \( a = \dfrac{0 - 25}{10} = -2.5 \) m/s².
\( F = ma = 500 \times (-2.5) = -1250 \) N. The brakes exert a force of 1250 N opposite to the motion.
In simple words: Find the retardation first and then multiply by the mass.

Teacher's Note:
1. The negative sign shows that the force opposes the motion.
2. Convert km/h to m/s by dividing by 3.6.

Question 7. A mass of 10 kg moving with a velocity of 10 m/s collides with a mass of 6 kg at rest. They stick together and move with a common velocity after the collision. Find the common velocity.
Answer:
Total momentum before = \( 10 \times 10 + 6 \times 0 = 100 \) kg m/s.
After the collision: \( (10 + 6)v = 100 \), so v = 6.25 m/s.
In simple words: The momentum is shared by the larger combined mass, so the speed drops.

Teacher's Note:
1. Momentum is conserved in all collisions.
2. Kinetic energy is not conserved when bodies stick together.

Question 8. Two masses of 1 g and 4 g are moving with equal kinetic energies. What is the ratio of the magnitudes of their linear momenta?
Answer:
Equal kinetic energies: \( \tfrac12 m_1v_1^2 = \tfrac12 m_2v_2^2 \), so \( \dfrac{v_1}{v_2} = \sqrt{\dfrac{m_2}{m_1}} = 2 \).
\( \dfrac{p_1}{p_2} = \dfrac{m_1v_1}{m_2v_2} = \dfrac{1 \times 2}{4} = \dfrac12 \). The ratio is 1 : 2.
In simple words: The lighter body moves twice as fast, but its momentum is still half.

Teacher's Note:
1. p = sqrt(2mK).
2. Compare speeds first, then momenta.