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All questions of Force and Acceleration for Grade 8 Exam

A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 12 m/s . If the mass of the ball is 0.15 kg, the impulse imported on the ball is
  • a)
    4.2 Ns
  • b)
    3.6 Ns
  • c)
    2.8 Ns
  • d)
    1.6 Ns
Correct answer is option 'B'. Can you explain this answer?

Mihir Shah answered
Understanding Impulse
Impulse is defined as the change in momentum of an object when a force is applied over a period of time. It can also be calculated using the formula:
Impulse = Change in Momentum = Final Momentum - Initial Momentum
Given Data
- Initial Speed of Ball (u): 12 m/s (towards the bowler)
- Final Speed of Ball (v): -12 m/s (back towards the batsman)
- Mass of Ball (m): 0.15 kg
Calculating Momentum
- Initial Momentum (p_initial):
p_initial = m * u = 0.15 kg * 12 m/s = 1.8 kg·m/s
- Final Momentum (p_final):
p_final = m * v = 0.15 kg * (-12 m/s) = -1.8 kg·m/s
Finding Change in Momentum
- Change in Momentum (Δp):
Δp = p_final - p_initial
= (-1.8 kg·m/s) - (1.8 kg·m/s)
= -3.6 kg·m/s
Calculating Impulse
Since impulse is equal to the change in momentum:
Impulse = Δp = -3.6 kg·m/s
The negative sign indicates a change in direction, but impulse is typically expressed as a positive value in magnitude.
Conclusion
Thus, the impulse imparted on the ball is 3.6 Ns, which corresponds to option B. This value represents the total effect of the force acting on the ball during the collision with the bat.

A bullet of mass 20 g is horizontally fired with a velocity of 150 ms–2 from a pistol of mass 2 kg. What is the recoil velocity of the pistol?
  • a)
    –1.25 ms–1
  • b)
    –1.5 ms–1
  • c)
    1.5 ms–1
  • d)
    2.5 ms–1
Correct answer is option 'B'. Can you explain this answer?

Recoil Velocity Calculation:

Given:
Mass of bullet (m₁) = 20 g = 0.02 kg
Velocity of bullet (v₁) = 150 m/s
Mass of pistol (m₂) = 2 kg

In this problem, we need to find the recoil velocity of the pistol after firing the bullet.

Using the principle of conservation of momentum, we can solve this problem.

1. Conservation of Momentum:
According to the principle of conservation of momentum, the total momentum before an event is equal to the total momentum after the event, as long as no external forces are acting on the system.

Mathematically, we can express this principle as:
m₁v₁ + m₂v₂ = m₁u₁ + m₂u₂

Where:
m₁ = mass of bullet
v₁ = initial velocity of bullet
m₂ = mass of pistol
v₂ = initial velocity of pistol (recoil velocity)
u₁ = final velocity of bullet (after firing)
u₂ = final velocity of pistol (recoil velocity)

2. Initial and Final Velocities:
Before firing the bullet, the pistol and the bullet are at rest. So, the initial velocities of both the bullet and the pistol are zero.

Therefore, the equation becomes:
(0.02 kg)(150 m/s) + (2 kg)(0) = (0.02 kg)(u₁) + (2 kg)(u₂)

3. Solving for Recoil Velocity:
Simplifying the equation, we get:
3 = 0.02u₁ + 2u₂

Since the bullet is fired horizontally, the final velocity of the bullet (u₁) is in the opposite direction to the initial velocity (v₁). Therefore, u₁ = -150 m/s.

Substituting the values into the equation, we get:
3 = (0.02)(-150) + 2u₂
3 = -3 + 2u₂
6 = 2u₂
u₂ = 6/2 = 3 m/s

Therefore, the recoil velocity of the pistol is 3 m/s in the opposite direction to the bullet's initial velocity.

A girl of mass 20 kg having velocity 2 m/sec jumps on stationary cart of mass 2 kg. Find the velocity of the girl when the cart starts moving
  • a)
    2 m/s
  • b)
    54 m/s
  • c)
    0.95 m/s
  • d)
    1.81 m/s
Correct answer is option 'D'. Can you explain this answer?

Advait Desai answered
To solve this problem, we can use the principle of conservation of momentum. This principle states that the total momentum before an event is equal to the total momentum after the event, as long as no external forces are acting on the system.

Let's break down the problem into two parts: before the girl jumps onto the cart and after she jumps onto the cart.

Before the girl jumps:
- Girl's mass (m1) = 20 kg
- Girl's velocity (v1) = 2 m/s
- Cart's mass (m2) = 2 kg (stationary, so its velocity is 0)

After the girl jumps:
- Girl's mass (m1) = 20 kg
- Girl's velocity (v1') = ?
- Cart's mass (m2) = 2 kg
- Cart's velocity (v2') = ?

Using the principle of conservation of momentum, we can write the equation:
(m1 * v1) + (m2 * 0) = (m1 * v1') + (m2 * v2')

Simplifying the equation, we get:
(m1 * v1) = (m1 * v1') + (m2 * v2')

Plugging in the values, we have:
(20 kg * 2 m/s) = (20 kg * v1') + (2 kg * v2')

Simplifying further, we have:
40 kg m/s = 20 kg * v1' + 2 kg * v2'

Since the cart starts moving, its velocity (v2') is not zero. Therefore, we can rewrite the equation as:
40 kg m/s = 20 kg * v1' + 2 kg * v2'

To find the velocity of the girl when the cart starts moving (v1'), we need to know the value of v2'. Unfortunately, the problem does not provide this information.

Therefore, we cannot determine the exact velocity of the girl when the cart starts moving. The correct answer should be option 'Cannot be determined' or 'Insufficient information provided'. The given options do not include this choice, so none of them is correct.

A force of 5 N gives a mass M1, an acceleration equal to 8 ms–2 and m2 an acceleration of 24 ms–2 What is the acceleration if both the masses are tied together?
  • a)
    3 ms–2
  • b)
    4 ms–2
  • c)
    6 ms–2
  • d)
    12 ms–2
Correct answer is option 'C'. Can you explain this answer?

Aarya Chavan answered
To solve this problem, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. In equation form, this is expressed as F = m * a.

Let's analyze the given information step by step:

1. Force acting on mass M1: The force acting on mass M1 is 5 N, and its acceleration is given as 8 m/s^2. Using Newton's second law, we can calculate the mass M1 as follows:
F = m1 * a1
5 N = m1 * 8 m/s^2
m1 = 5 N / 8 m/s^2
m1 = 0.625 kg

2. Force acting on mass M2: The force acting on mass M2 is also 5 N, but its acceleration is given as 24 m/s^2. Using Newton's second law, we can calculate the mass M2 as follows:
F = m2 * a2
5 N = m2 * 24 m/s^2
m2 = 5 N / 24 m/s^2
m2 = 0.208 kg

3. Combined force acting on both masses: When the masses M1 and M2 are tied together, the force acting on them remains the same, which is 5 N. The total mass of the combined system is the sum of the individual masses, given by:
Total mass = m1 + m2
Total mass = 0.625 kg + 0.208 kg
Total mass = 0.833 kg

Now, we can find the acceleration of the combined system by rearranging Newton's second law equation:
F = m * a
5 N = 0.833 kg * a
a = 5 N / 0.833 kg
a ≈ 6 m/s^2

Therefore, the acceleration when both masses are tied together is approximately 6 m/s^2. Hence, the correct answer is option 'C'.

A constant force acts on an object of mass 5 kg for a duration of 2 seconds. It increases the object’s velocity from 3 m/s to 7m/s, find the magnitude of the force applied on the object.
  • a)
    10 N
  • b)
    20 N
  • c)
    25 N
  • d)
    30 N
Correct answer is option 'A'. Can you explain this answer?

Given data:
- Mass of the object (m) = 5 kg
- Initial velocity (u) = 3 m/s
- Final velocity (v) = 7 m/s
- Time duration (t) = 2 seconds

Calculation:
- Using the formula:
\[ F = \frac{m(v-u)}{t} \]
- Substituting the given values:
\[ F = \frac{5(7-3)}{2} = \frac{5 \times 4}{2} = \frac{20}{2} = 10 \, N \]
Therefore, the magnitude of the force applied on the object is 10 N. Hence, option 'A' is the correct answer.

A bullet of mass 20 g is horizontally fired with a velocity of 150 ms–2 from a pistol of mass 2 kg.
What is the recoil velocity of the pistol?
  • a)
    – 1.25 ms–1
  • b)
    – 1.5 ms–1
  • c)
    1.5 ms–1
  • d)
    2.5 ms–1
Correct answer is option 'B'. Can you explain this answer?

M1 = 20 g = 0.02 kg, m2 = 2 kg.
u1 = 0, u2 = 0 (both pistol and bullet are at rest)
v1 = + 150 ms–1 (the find velocity of the bullet)
let v2 be the recoil velocity of the pistol
according to the law of conservation, total
moment remains the same.
m1u1 + m2u2 = m1v1 + m2v2
.02 × 0 + 2 × 0 = 0.02 × 150 + 2 × v2
0 = 3 + 2v2
⇒ v2 = -3/2 = – 1.5 ms–1
Negative sign indicates that the pistol will recoil in the opposite direction of the bullet.

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