Numerical Problems (Solved): Force and Newton's Law of Motion | Science Class 9 PDF Download

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Q1: A cricket ball of mass 70 g moving with a velocity of 0.5 m s^-1 is stopped by player in 0.5 s. What is the force applied by player to stop the ball?  
Sol. Here m = 70 g = 0.070 kg; u = 0.5 m s^-1; v = 0; t = 0.5 s 

Q2: What will be acceleration of a body of mass 5 kg if a force of 200 N is applied to it?  
Sol. Here m = 5 kg; F = 200 N 
F = ma or a = F/m 
 

Q3: A bullet of mass 10 g is fired from a rifle. The bullet takes 0.003 s to move through its barrel and leaves with a velocity of 300 ms^-1. What is the force exerted on the bullet by the rifle?  
Sol. Here m = 10 g = 0.010 kg ; u = a ; v = 300 m s^-1 
t = 0.003 s, F = ? 

Q4: A truck of mass 1500 kg is moving along a straight road with a uniform velocity of 72 km/h. Its velocity is reduced to 36 km/h in 6 seconds due to an external braking force. Calculate the acceleration and the change in momentum. Also calculate the magnitude of the force required.
Sol. Mass of the truck: m = 1500 kg
Initial velocity of the truck: u = 72 km/h

u = 72 × 518 m/s = 20 m/s
Final velocity of the truck: v = 36 km/h
v = 36 × 518 m/s = 10 m/s
Time: t = 6 s
Acceleration:
a = v - ut = 10 - 206 = -106 m/s² = -1.67 m/s²
Momentum change:
Δp = mv - mu or m(v - u)
= 1500 × (10 - 20) = 1500 × (-10) = -15000 kg·m/s
Magnitude of force: F = ma
= 1500 × (-1.67) = -2505 N
The acceleration, momentum change, and force are opposing the motion of the truck, as indicated by the negative sign.

Q5: A cricket ball of mass 200 g is moving with a speed of 40 m/s and is caught by a fielder, bringing it to rest in 0.02 s. What is the force exerted by the fielder's hand on the ball?
Sol. Mass of the ball: M = 200 g = 0.2 kg
Initial velocity of the ball: u = 40 m/s
Final velocity of the ball: v = 0
Time: t = 0.02 s
Acceleration of the ball:
a = v - ut = 0 - 400.02 = -2000 m/s²
Force exerted by the fielder's hand: F = ma
= (0.2) × (-2000 m/s²) = -400 N (opposite to the motion of the ball).

Q6: How much momentum will a basketball of mass 5 kg transfer to the ground if it falls from a height of 50 cm and does not rebound? Take its downward acceleration to be 10 m/s2.
Sol. Height: h = 50 cm = 0.5 m
Mass: m = 5 kg
Initial velocity: u = 0
Acceleration: a = 10 m/s²
Final velocity:
v = √(u² + 2ah) = √(0 + 2 × 10 × 0.5) = √10 m/s = 3.16 m/s
Momentum transferred:
mv = 5 kg × 3.16 m/s = 15.8 kgm/s

Q7: Which would require greater force: accelerating a 10 g mass at 5 m s^_2 or 20 g mass at 2 m s^-2?  
Sol. In first case m₁ = 10 g = kg = 0.010 kg; 
Now a₁ = 5 ms^-2 ; F₁ = ? 
F₁ = m₁a₁ = 0.010 × 5
F₁ = 0.050 newton 
In second case, m₂ = 20 g =0.020 kg 
or m₂ = 0.020 kg 
a₂ = 2 m s^-2 ; F₂ = ? 
Now F₂ = m₂a₂ = 0.020 × 2 
or F₂ = 0.04 newton 
We find that F₁ > F₂, hence more force is required to accelerate 10 g at 5 m s^-2 than accelerating 20 g at 2 ms^-2.  

Q8: A force of 5 N gives a mass m₁, an acceleration of 8 ms^-2 and a mass m₂, an acceleration of 24 m s^-2. What acceleration would it give if both the masses are tied together?  
Sol. Let us first find mass m₁ and m₂. 
F = m₁ a₁ 
5 =m₁ (8) or m₁ = 5/8 kg 
F = m₂ a₂ 
5 = m₂ (24) or m₂ = 5/24 kg 
Total mass M = m₁ + m₂ 

Let A be the acceleration produced in mass M. 
F = MA 

Hence the acceleration of the combination is 6 ms^-2. 

Q9: A cart of mass 150 kg is accelerated uniformly from a velocity of 4 m/s to 10 m/s in 5 s. Calculate the initial and final momentum of the cart. Also, find the magnitude of the force exerted on the cart.

Sol. Mass of the cart: m = 150 kg
Initial velocity: u = 4 m/s
Final velocity: v = 10 m/s
Time: t = 5 s
Initial momentum:
P_initial = mu = 150 × 4 = 600 Ns
Final momentum:
P_final = mv = 150 × 10 = 1500 Ns
Force exerted on the cart:
F = mv - mut = 1500 - 6005 = 9005 = 180 N

Q10: A car of mass 2 kg traveling in a straight line with a velocity of 15 m/s collides with and sticks to a stationary cart of mass 8 kg. They then move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.

Sol. Mass of the car: m₁ = 2 kg
Velocity of the car before collision: v₁ = 15 m/s
Mass of the stationary cart: m₂ = 8 kg
Velocity of the cart before collision: v₂ = 0
Total momentum before collision:
P_before = m₁v₁ + m₂v₂ = (2 × 15) + (8 × 0) = 30 kg·m/s
It is given that after the collision, the car and the cart stick together.
Total mass of the combined system: (m₁ + m₂)
Velocity of the combined object: v
According to the law of conservation of momentum:
Total momentum before collision = Total momentum after collision
(m₁v₁ + m₂v₂) = (m₁ + m₂) v
(2 × 15) + (8 × 0) = (2 + 8) v
30 = 10v
v = 3010 = 3 m/s
Total momentum just before the impact: 30 kgm/s
Total momentum just after the impact: (m₁ + m₂) v = 10 × 3 = 30 kgm/s
Hence, the velocity of the combined object after collision = 3 m/s.