State the principle of conservation of linear momentum. [2] (b) A stationery ball of mass, 6.0x10 kg is hit horizontally with a tennis racket. The ball is in contact with the racket for 30 ms and leaves the racket with a speed of 27 ms. Calculate the change in momentum of the ball, 2. average force which the racket exerts on the ball. 3. horizontal distance travelled by the ball before it hits the ground if it leaves the racket at a vertical height of 2.5 m (iii) Suggest a reason why the collision between the ball and the racket is
(b) The change in momentum of the ball can be calculated using the formula:
Change in momentum = mass of the ball × final velocity - mass of the ball × initial velocity
Given:
Mass of the ball (m) = 6.0 × 10 kg
Initial velocity (u) = 0 m/s (since the ball is stationary)
Final velocity (v) = 27 m/s
Change in momentum = 6.0 × 10 kg × 27 m/s - 6.0 × 10 kg × 0 m/s
Change in momentum = 1620 kg·m/s
Therefore, the change in momentum of the ball is 1620 kg·m/s.
To calculate the average force exerted by the racket on the ball, we can use the formula:
Average force = Change in momentum / Time
Given:
Change in momentum = 1620 kg·m/s
Time (t) = 30 ms = 0.03 s
Average force = 1620 kg·m/s / 0.03 s
Average force ≈ 54,000 N
Therefore, the average force exerted by the racket on the ball is approximately 54,000 N.
To calculate the horizontal distance travelled by the ball before it hits the ground, we assume that air resistance and other external factors are negligible.
Using the equation of motion for horizontal motion:
Distance (d) = Horizontal velocity (v) × Time (t)
Given:
Horizontal velocity (v) = 27 m/s
Time (t) = ?
We need to find the time taken by the ball to hit the ground at a vertical height of 2.5 m. We can use the equation of motion for vertical motion:
Vertical distance (h) = Initial vertical velocity (u) × Time (t) + 0.5 × Acceleration due to gravity (g) × (Time (t))^2
Given:
Vertical distance (h) = 2.5 m
Initial vertical velocity (u) = 0 m/s (since the ball leaves the racket horizontally)
Acceleration due to gravity (g) = 9.8 m/s^2
2.5 = 0.5 × 9.8 × (Time (t))^2
(Time (t))^2 = 2.5 / (0.5 × 9.8)
(Time (t))^2 = 0.2551
Time (t) ≈ √0.2551
Time (t) ≈ 0.5054 s
Now that we have found the time for the ball to hit the ground, we can calculate the horizontal distance travelled using the equation:
Distance (d) = Horizontal velocity (v) × Time (t)
Distance (d) = 27 m/s × 0.5054 s
Distance (d) ≈ 13.6478 m
Therefore, the horizontal distance travelled by the ball before it hits the ground is approximately 13.6478 m.
(iii) One reason why the collision between the ball and the racket is elastic is that there is no mention of any energy loss during the collision. In an elastic collision, the total kinetic energy remains constant before and after the collision. If there were mentions of energy lost, such as due to friction or deformation of the ball, then the collision would be inelastic.
If there is no external force on a system, its momentum vector does not change.
I do not believe this tennis ball is 60 = 6.0 * 10 kilograms. That makes the rest crazy.
My apologies for the confusion caused by the erroneous value for the mass of the tennis ball. Let's correct the calculations.
Given:
Mass of the ball (m) = 0.060 kg (assumed value)
Initial velocity (u) = 0 m/s (since the ball is stationary)
Final velocity (v) = 27 m/s
Time (t) = 30 ms = 0.030 s
Vertical distance (h) = 2.5 m
Acceleration due to gravity (g) = 9.8 m/s^2
Now let's calculate:
Change in momentum = mass of the ball × final velocity - mass of the ball × initial velocity
Change in momentum = 0.060 kg × 27 m/s - 0.060 kg × 0 m/s
Change in momentum = 1.62 kg·m/s
Average force = Change in momentum / Time
Average force = 1.62 kg·m/s / 0.030 s
Average force ≈ 54 N
Use the equation of motion for vertical motion:
Vertical distance (h) = Initial vertical velocity (u) × Time (t) + 0.5 × Acceleration due to gravity (g) × (Time (t))^2
2.5 m = 0 × 0.030 s + 0.5 × 9.8 m/s^2 × (0.030 s)^2
2.5 m = 0 + 0.5 × 9.8 m/s^2 × 0.0009 s^2
2.5 m = 0.00441 m
0.00441 m = 2.5 m
This equation is not valid. I apologize for the mistake in my previous response. Since the ball leaves the racket horizontally, its vertical distance won't be affected by the horizontal motion. The vertical distance travelled by the ball before it hits the ground cannot be determined with the given information.
Let me know if there is anything else I can assist you with.
The principle of conservation of linear momentum states that the total momentum of a system of objects remains constant if no external forces act on the system.
To calculate the change in momentum of the ball, we can use the formula:
Change in momentum = final momentum - initial momentum
Given:
Mass of the ball (m) = 6.0x10 kg
Initial velocity (u) = 0 m/s (since the ball is stationary)
Final velocity (v) = 27 m/s
Initial momentum = mass x initial velocity
Initial momentum = 6.0x10 kg x 0 m/s = 0 kg·m/s
Final momentum = mass x final velocity
Final momentum = 6.0x10 kg x 27 m/s = 1620 kg·m/s
Change in momentum = 1620 kg·m/s - 0 kg·m/s = 1620 kg·m/s
Therefore, the change in momentum of the ball is 1620 kg·m/s.
To calculate the average force exerted by the racket on the ball, we can use the formula:
Average force = change in momentum / time
Given:
Change in momentum = 1620 kg·m/s
Time (Δt) = 30 ms = 0.03 s
Average force = 1620 kg·m/s / 0.03 s = 54,000 N
Therefore, the average force exerted by the racket on the ball is 54,000 N.
To calculate the horizontal distance traveled by the ball before it hits the ground, we can use the formula:
Distance (d) = horizontal velocity x time
Given:
Final velocity (v) = 27 m/s
Time (Δt) = ?
Vertical height (h) = 2.5 m
Since the ball leaves the racket horizontally, the vertical motion is independent of the horizontal motion. We can use the kinematic equation:
h = (1/2)gt^2
Solving for time (Δt):
2.5 m = (1/2)(9.8 m/s^2)(Δt)^2
Δt^2 = (2.5 m x 2) / 9.8 m/s^2
Δt^2 = 0.5102 s^2
Δt ≈ 0.71 s
Since the time of flight is equal to the time of contact between the ball and racket, we can use this as the value for Δt.
Using the formula for distance:
Distance (d) = horizontal velocity x time
Distance (d) = 27 m/s x 0.71 s ≈ 19.17 m
Therefore, the horizontal distance traveled by the ball before it hits the ground is approximately 19.17 m.
Reason for the collision between the ball and the racket being elastic:
The collision between the ball and the racket is likely to be elastic because the ball leaves the racket with a significant speed, which indicates that there was no significant loss of kinetic energy during the collision. Additionally, an elastic collision would satisfy the principle of conservation of linear momentum, as stated earlier.