A ball of mass 0.2 kg strikes a force sensor and sticks to it. Just before impact the ball is travelling horizontally at a speed of 4.0 m s–1. The graph shows the variation with time t of the force F recorded by the sensor.

Question

Create an image that visualizes a physics problem. The image should depict a small ball with a mass of 0.2 kg moving horizontally at a speed of 4.0 m/s towards a force sensor. Please depict the moment of impact when the ball sticks to the sensor. Include the graph in the image as well. The graph has two constant lines, one upward and one downward, representing the variation with time. The upward line stops at 400 ms and the downward one starts at 440 ms. Please ensure the image does not contain any text.

A ball of mass 0.2 kg strikes a force sensor and sticks to it. Just before impact the ball is travelling horizontally at a speed of 4.0 m s–1. The graph shows the variation with time t of the force F recorded by the sensor.

(plus there is this graph where there is this constant line upward then constant line downwards, the upwards line has a t/ms of 400 and the line downwards has 440)
HEEELPPP!

What Is maximum force?

Answer 1

m = 0.2 kg, u = 4 m/s, v = 0, t = 40 ms = 0.02 s (see x- axis)

F = (mv - mu)/ t
F = (0.2 x 0 - 0.2x4)/0.02
F = 40 N

👍 1 👎 0

Answer 2

the other responses didn't make sense to me, so here's an answer that made more sense to me:

p (momentum) = mv
F = Δ p/Δ t

p right before the ball hits the force sensor is p=(0.2kg x 4m/s) = 0.8m/s
p at Fmax would be 0, because there is 0 velocity
t right before the ball hits the force senor is 400ms = 0.40s
t at Fmax is 420ms = 0.42s

So now we can solve for F:
F= Δ p/Δ t
F= 0.8kg m/s - 0.0kg m/s / 0.42s - 0.40s
F= 0.8kg m/s / 0.02s
F= 40 kg m/s^2
So, 40N would be the answer.

Answer 3

To find F_max, we can utilize the impulse-momentum theorem. The impulse experienced by the ball during the collision is equal to the change in momentum of the ball. The impulse can also be represented as the product of the average force and the time duration of the collision. Therefore, we have:

Impulse = change in momentum = average force × time duration

Since the ball sticks to the force sensor, its final velocity is 0 m/s. The initial momentum of the ball can be calculated using its mass and initial velocity:

Initial momentum = mass × initial velocity = 0.2 kg × 4.0 m/s = 0.8 kg m/s

The change in momentum is equal to the initial momentum (since the final momentum is 0):

Change in momentum = 0.8 kg m/s

The time duration of the collision is given as the base of the triangle in the graph, which is 40 ms (440 ms - 400 ms = 40 ms). Now, we can find the average force during the collision:

Average force = change in momentum / time duration = 0.8 kg m/s / 0.04 s = 20 N

Since the graph is a triangle, the peak force (F_max) will be twice the average force:

F_max = 2 × average force = 2 × 20 N = 40 N

Therefore, F_max is 40 N.

Answer 4

This matches with the previous answer that utilized the impulse-momentum theorem.

Answer 5

sorry, when i said "p right before the ball hits the force sensor is p=(0.2kg x 4m/s) = 0.8m/s" it should actually be 0.8kg m/s

Answer 6

Look at the area of the triangle

Answer 7

m = 0.2 kg, u = 4 m/s, v = 0, t = 40 ms = 0.04 s (see x- axis)

F = (mv - mu)/ t
F = (0.2 x 0 - 0.2x4)/0.04
F = 20 N