Impulse – problems and solutions

1. A 0.5 kg ball free fall from a height of h_{1} = 7.2 meters and reflected a height of h_{2} = 3.2 meters. Acceleration due to gravity = 10 m/s^{2}. Determine impulse.

__Known :__

Mass of ball (m) = 0.5 kg

First height (h_{1}) = 7.2 meter

Second height (h_{2}) = 3.2 meter

Acceleration due to gravity (g) = 10 m/s^{2}

__Wanted :__ Impulse (I)

__Solution :__

__Velocity of ball before collisions ____(v___{o}__)__

Velocity of ball before collision calculated using equation of free fall motion. __Known :__ Height (h) = 7.2 meters, acceleration due to gravity (g) = 10 m/s^{2}. __Wanted :__ Final velocity after collision.

v^{2} = 2 g h

v_{o}^{2} = 2(10)(7.2) = 144

v_{o} = 2(10)(7.2) = 12 m/s

Velocity of ball before collision (v_{o}) = -12 m/s. Minus sign indicates the direction of ball.

__Velocity of ball after collision ____(v___{t}__)__

Velocity of ball after collision calculated using equation of vertical motion. __Known __: height (h) = 3.2 meters, acceleration due to gravity (g) = -10 m/s^{2}, final velocity at the maximum height (v_{t}^{2}) = 0. __Wanted :__ Initial velocity after collision between ball and floor.

v_{t}^{2} = v_{o}^{2} + 2 g h

0 = v_{o}^{2} + 2 (-10)(3.2)

v_{o}^{2} = 64

v_{o} = √64 = 8 m/s

Velocity of ball after collision (v_{t}) is 8 m/s

__Impulse (I)__

Impulse (I) = the change in momentum (Δp)

I = m (v_{t }– v_{o}) = (0.5)(8-(-12)) = (0.5)(8 + 12) = (0.5)(20) = 10 Newton second

2. A 5-gram ball free fall from a height and strikes the floor. Acceleration due to gravity, g = 10 ms^{-2}. Velocity of ball before collision is 6 ms^{-1} and after collision, the ball is reflected upright at 4 m/s. Determine the impulse.

__Known :__

Mass of ball (m) = 5 gram = 0.005 kg

Velocity of ball before collision (v_{o}) = -6 m/s

Velocity of ball after collision (v_{t}) = 4 m/s

Plus and minus sign indicates that the direction before collision is opposite with the direction after collision.

__Wanted:__ Impulse (I)

__Solution :__

Impulse (I) = the total change in momentum (Δp).

I = Δp = m v_{t} – m v_{o }= m (v_{t} – v_{o})

I = (0.005)(4 – (-6)) = (0.005)(4 + 6) = (0.005)(10) = 0.05 Newton second

3. A 20-gram thrown with velocity of 4 m.s^{-1} to the left. After colliding with the wall, the ball is reflected with velocity of v_{2 }= 2 m.s^{-1} to the right. Determine the impulse.

__Known :__

Mass of ball (m) = 20 gram = 0.020 kg

Velocity of ball before collision (v_{o}) = -4 m/s (to the left)

Velocity of ball after collision (v_{t}) = +2 m/s (to the right)

Plus and minus sign indicates the opposite direction.

__Wanted :__ Impulse

__Solution :__

Impulse (I) = the change in momentum (Δp) = m v_{t} – m v_{o}

Impulse (I) = m (v_{t} – v_{o}) = 0.02 (2 – (-4))

Impulse (I) = 0.02 (2 + 4) = 0.02 (6)

Impulse (I) = 0.12 Newton second.

4. A 2 kg ball moving at 5 m/s strikes a wall and bounces back with a velocity of -5 m/s. Find the impulse exerted on the ball.

Solution: Impulse = change in momentum = m(v₂ – v₁) = 2(-5 – 5) = -20 kg·m/s

5. A 3 kg object experiences a force of 6 N for 4 seconds. What is the impulse experienced by the object?

Solution: Impulse = FΔt = 6 × 4 = 24 N·s

6. A 0.5 kg object is at rest and then accelerated by an impulse of 10 N·s. What is the final velocity?

Solution: Final velocity = Impulse / mass = 10 / 0.5 = 20 m/s

7. An impulse of 30 N·s acts on a 6 kg object. What is the change in velocity?

Solution: Δv = Impulse / mass = 30 / 6 = 5 m/s

8. A 100 g ball moving at 3 m/s is stopped by an impulse. What is the value of the impulse?

Solution: Impulse = m(v₂ – v₁) = 0.1(0 – 3) = -0.3 kg·m/s

9. A 2 kg object experiences a 4 N force for 5 seconds. Find the impulse.

Solution: Impulse = FΔt = 4 × 5 = 20 N·s

10. An object of mass 3 kg is moving at 3 m/s and comes to rest due to an impulse. Find the impulse.

Solution: Impulse = m(v₂ – v₁) = 3(0 – 3) = -9 kg·m/s

11. A 5 kg object experiences a force of 2 N for 3 seconds. What is the impulse?

Solution: Impulse = FΔt = 2 × 3 = 6 N·s

12. An impulse of 8 N·s acts on a 4 kg object. What is the change in velocity?

Solution: Δv = Impulse / mass = 8 / 4 = 2 m/s

13. A 1 kg ball moving at 2 m/s strikes a wall and reverses its direction at the same speed. Find the impulse exerted on the ball.

Solution: Impulse = m(v₂ – v₁) = 1(-2 – 2) = -4 kg·m/s

14. A 4 kg object experiences a 3 N force for 3 seconds. Find the impulse.

Solution: Impulse = FΔt = 3 × 3 = 9 N·s

15. An impulse of 10 N·s acts on a 2 kg object. What is the change in velocity?

Solution: Δv = Impulse / mass = 10 / 2 = 5 m/s

16. A 3 kg object experiences a force of 5 N for 2 seconds. What is the impulse experienced by the object?

Solution: Impulse = FΔt = 5 × 2 = 10 N·s

17. An object of mass 4 kg is moving at 5 m/s and comes to rest due to an impulse. Find the impulse.

Solution: Impulse = m(v₂ – v₁) = 4(0 – 5) = -20 kg·m/s

18. A 2 kg object experiences a 6 N force for 3 seconds. Find the impulse.

Solution: Impulse = FΔt = 6 × 3 = 18 N·s

19. An impulse of 12 N·s acts on a 3 kg object. What is the change in velocity?

Solution: Δv = Impulse / mass = 12 / 3 = 4 m/s

20. A 5 kg object experiences a force of 1 N for 4 seconds. What is the impulse?

Solution: Impulse = FΔt = 1 × 4 = 4 N·s

21. An impulse of 15 N·s acts on a 5 kg object. What is the change in velocity?

Solution: Δv = Impulse / mass = 15 / 5 = 3 m/s

22. A 3 kg ball moving at 4 m/s is stopped by an impulse. What is the value of the impulse?

Solution: Impulse = m(v₂ – v₁) = 3(0 – 4) = -12 kg·m/s

23. A 2 kg object experiences a 5 N force for 1 second. Find the impulse.

Solution: Impulse = FΔt = 5 × 1 = 5 N·s

**What is impulse?****Answer:**Impulse is the product of a force and the time interval over which it acts on an object. It represents the change in momentum of the object and is given by the formula $Impulse=F×Δt$.

**How does impulse relate to momentum?****Answer:**Impulse is equal to the change in momentum of an object. If an object experiences an impulse, its momentum will change by that same amount.

**Why is it safer to land on a soft mat than a hard floor when jumping from a height?****Answer:**A soft mat increases the time taken to bring the jumper to a stop, compared to a hard floor. A longer stopping time means a smaller average force exerted on the jumper, reducing the risk of injury. This longer time results in a reduced force but the same impulse.

**How do airbags in cars work in terms of impulse?****Answer:**Airbags inflate rapidly during a collision, increasing the time over which a person’s momentum is brought to zero. By increasing the time interval of the force (deceleration), the average force exerted on the person is reduced, which decreases the risk of injury.

**Why do baseball players “follow through” when hitting a ball?****Answer:**Following through increases the contact time between the bat and the ball. A longer contact time means a greater impulse can be applied to the ball, which can increase the change in the ball’s momentum, potentially resulting in a harder hit.

**Why are goods in transport often packed with cushioning materials?****Answer:**Cushioning materials, like foam or bubble wrap, increase the time taken for an object to come to a stop when it experiences a force (e.g., during a sudden stop or a drop). This reduces the average force on the object for a given impulse, helping to prevent damage.

**How does impulse explain the effect of bouncing a basketball with more force?****Answer:**Applying more force when bouncing a basketball increases the impulse imparted to the ball, resulting in a greater change in the ball’s momentum. This means the ball will rebound with a higher velocity.

**In terms of impulse, why are athletes advised to “roll” when they fall?****Answer:**Rolling when falling increases the time over which the change in momentum happens. This means that the average force exerted on the athlete’s body is reduced, decreasing the potential for injury.

**If two forces of different magnitudes act on an object for the same duration, how will their impulses compare?****Answer:**The impulse is the product of force and time. Since the time duration is the same for both forces, the impulse of the larger force will be greater than that of the smaller force.

**Why do golfers use a “swinging” motion when hitting a golf ball, in terms of impulse?**

**Answer:**The swinging motion increases the time of contact between the golf club and the ball. A longer contact time allows for a greater impulse to be delivered to the ball, leading to a larger change in the ball’s momentum and, thus, a farther drive.