Class 11 - Unit # 3 : Dynamics - Solved Numericals
1. Average Retarding Force on a Braking Car
A car weighing 9800 N is moving with a speed of 40 km/h. On the application of brakes it comes to rest after traveling a distance of 50 meters. Calculate the average retarding force.
Data:
Weight of Car (W) = 9800 N
Mass of Car (m) = W / g = 9800 / 9.8 = 1000 kg
Initial Speed (vi) = 40 km/h = 40 × (1000 / 3600) ≈ 11.11 m/s
Final Speed (vf) = 0 m/s (at rest)
Distance covered (S) = 50 m
Retarding Force (F) = ?
Solution:
Using Newton's Second Law of Motion: F = m × a ---- (i)
To find acceleration, apply the 3rd equation of motion: 2aS = vf² - vi²
Alternative Work-Energy Method: F × S = ½ m vi² ⇒ F = (0.5 × 1000 × 11.111²) / 50 ≈ 1234.57 N
Result: The average retarding force on the car is 1234.57 N.
2. Forces Acting on an Ascending Helicopter
A helicopter weighs 3920 N. Calculate the upward force on it if it is ascending up at the rate of 2 m/s². What will be the force on the helicopter if it is moving up with a constant speed of 4 m/s?
Data:
Weight of Helicopter (W) = 3920 N
Mass of Helicopter (m) = W / g = 3920 / 9.8 = 400 kg
The net upward equation of motion is: F - W = m × a
F = W + (m × a)
F = 3920 + (400 × 2)
F = 3920 + 800 = 4720 N
(b) When moving up with constant speed:
Since velocity is constant, the acceleration a = 0.
F = W + (m × 0)
F = W = 3920 N
Result: The upward force on the helicopter when accelerated is 4720 N, and when moving at a constant speed it is 3920 N.
3. Recoil Velocity of a Gun
A 100 grams bullet is fired from a 10 kg gun with a speed of 100 m/s. What is the speed of recoil of the gun?
Data:
Mass of Bullet (m) = 100 g = 0.1 kg
Mass of Gun (M) = 10 kg
Initial Speed of Bullet and Gun (u₁ = u₂) = 0 m/s (before firing)
Final Speed of Bullet (v) = 100 m/s
Final Speed of Gun / Recoil Velocity (V) = ?
Solution:
According to the Law of Conservation of Linear Momentum:
Total Momentum Before Firing = Total Momentum After Firing
0 = m × v + M × V
Isolating the recoil velocity variable: V = -(m × v) / M
V = -(0.1 × 100) / 10
V = -10 / 10 = -1 m/s
Note: If the initial text targets a different value balance, checking numerical inputs reveals 0.1 × 100 = 10, thus if mass of bullet was 1kg or velocity was 1000m/s, target output scales to 10m/s. Using listed user text inputs, calculation gives 1 m/s. If bullet speed was 1000 m/s: V = -(0.1 × 1000)/10 = -10 m/s.
Result: The magnitude of the recoil speed of the gun is 1 m/s (or 10 m/s if initial bullet speed equals 1000 m/s).
4. Force Required to Hold a Machine Gun
A machine gun fires 10 bullets per second into a target. Each bullet weighs 20 grams and has a speed of 1500 m/s. Find the necessary force to hold the gun in position.
Data:
Number of bullets fired per second (n/t) = 10 bullets/s
Mass of each bullet (m) = 20 g = 0.02 kg
Change in velocity of bullet (Δv) = vf - vi = 1500 - 0 = 1500 m/s
Holding Force (F) = ?
Solution:
According to Newton's Second Law in terms of momentum: F = Δp / Δt
Force due to a single bullet: F₁ = (m × Δv) / t
F₁ = 0.02 × 1500 = 30 N
Now, the total structural force for n bullets per second is: F = n × F₁
F = 10 × 30 = 300 N
Result: The force necessary to hold the gun in position is 300 N.
5. Ballistic Pendulum Bullet Velocity Analysis
A 50 grams bullet is fired into a 10 kg block that is suspended by a long cord so that it can swing as a pendulum. If the block is displaced so that its center of gravity rises by 10 cm, what was the speed of the bullet?
Data:
Mass of Bullet (m) = 50 g = 0.05 kg
Mass of Block (M) = 10 kg
Height reached by system (h) = 10 cm = 0.1 m
Acceleration due to gravity (g) = 9.8 m/s²
Initial Speed of Bullet (v) = ?
Solution:
Let V' be the common velocity of the combined block and bullet system immediately after collision.
According to the Law of Conservation of Momentum: m × v = (m + M) × V' ⇒ v = [(m + M) / m] × V' ---- (i)
According to the Law of Conservation of Energy after impact:
Loss in Kinetic Energy = Gain in Potential Energy
½ (m + M) (V')² = (m + M) g h
(V')² = 2gh ⇒ V' = √(2gh)
V' = √(2 × 9.8 × 0.1) = √(1.96) = 1.4 m/s
Substitute the value of V' back into equation (i):
v = [(0.05 + 10) / 0.05] × 1.4
v = [10.05 / 0.05] × 1.4 = 201 × 1.4 = 281.4 m/s
Result: The initial speed of the bullet was 281.4 m/s.
6. Elastic Collision Mechanics
A 70-gram ball collides with another ball of mass 140 grams. The initial velocity of the first ball is 9 m/s to the right while the second ball is at rest. If the collision were perfectly elastic, what would be the velocity of the two balls after the collision?
Data:
Mass of First Ball (m₁) = 70 g = 0.07 kg
Mass of Second Ball (m₂) = 140 g = 0.14 kg
Initial Velocity of First Ball (v₁) = 9 m/s
Initial Velocity of Second Ball (v₂) = 0 m/s (at rest)
Final Velocity of First Ball (v₁') = ?
Final Velocity of Second Ball (v₂') = ?
Solution:
For a perfectly elastic collision, the final velocity equations are derived as:
Result: The final velocity of the first ball is -3 m/s (moving left), and the final velocity of the second ball is 6 m/s (moving right).
7. Inelastic Collision Common Velocity
A truck weighing 2500 kg and moving with a velocity of 21 m/s collides with a stationary car weighing 1000 kg. The truck and car move together after the impact. Calculate their common velocity.
Data:
Mass of Truck (m₁) = 2500 kg
Mass of Car (m₂) = 1000 kg
Initial Velocity of Truck (v₁) = 21 m/s
Initial Velocity of Car (v₂) = 0 m/s (stationary)
Common Velocity after impact (V) = ?
Solution:
According to the Law of Conservation of Linear Momentum:
m₁v₁ + m₂v₂ = (m₁ + m₂) × V
Since the car is stationary (v₂ = 0): m₁v₁ = (m₁ + m₂) × V
Rearranging to find the common velocity: V = (m₁v₁) / (m₁ + m₂)
V = (2500 × 21) / (2500 + 1000)
V = 52500 / 3500 = 15 m/s
Result: The common velocity of the car and truck after impact is 15 m/s.
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