A 0.1-kg object is traveling to the right (in the positive direction) with a speed of 2 m/s. After a 0.2 s collision, the object is traveling to the left at 3 m/s. What is the magnitude of the impulse (in N-s) acting on the object? The answer must be positive.
Mass x (velocity change) = 0.1 x 5 kg m/s
= 0.5 N s
(1 kg m/s is the same as 1 N*s)
The duration of the collision doesn't matter in oomputing the impulse, but it does determine the force involved.
a horizontal force of 100N is used to push a 20 kg box across a level floor for 10 m. how much work is done? how much work is needed to raise the same box by 10m?
20,000
20000
To determine the magnitude of the impulse acting on the object, we need to use the impulse-momentum principle. The impulse experienced by an object is equal to the change in momentum of the object.
We can calculate the change in momentum using the formula:
Δp = m * Δv
where Δp is the change in momentum, m is the mass of the object, and Δv is the change in velocity.
Given that the mass of the object is 0.1 kg, the initial velocity is 2 m/s to the right, and the final velocity is 3 m/s to the left, we can calculate the change in velocity:
Δv = final velocity - initial velocity
= (-3 m/s) - (2 m/s)
= -5 m/s
As the question specifically asks for the magnitude of the impulse, we can take the absolute value of the change in velocity:
|Δv| = |-5 m/s|
= 5 m/s
Now, we can calculate the impulse:
Impulse = m * Δv
= (0.1 kg) * (5 m/s)
= 0.5 kg * m/s
The magnitude of the impulse acting on the object is 0.5 N-s.
Note: The impulse is a vector quantity, meaning it has both magnitude and direction. However, in this question, we are asked for the magnitude only, which is always positive.