A0.40-kg object is traveling to the right (in the positive direction) with a speed of 4.0 m/s. after a 0.20 s collision, the object is traveling to the left at 3.0 m/s. what is the magnitude of the impulse (in n-s) acting on the object? the answer must be positive. (never include units in the answer to a numerical question.)

Explanation:

It is given that,

Mass of the object, m = 0.4 kg

Speed of the object, u = 4 m/s

After the collision, the object is travelling, v = -3 m/s (to left)

Time, t = 0.2 seconds

Let J is the impulse acting on the object. We know that the impulse of an object is equal to the change in momentum. It can be calculated as :

J = -2.8 kg-m/s

|J| = 2.8 kg-m/s

So, the impulse acting on the object has a magnitude of 2.8 kg-m/s. Hence, this is the required solution.

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this

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