The combine mass of two objects sticking together is 1500 kg and is traveling at 63 m/s. If object 1 has a mass of 400 kg and was traveling at 45 m/s. Then how fast was object 2 moving before the impact?

(a) -16.7 N s; (b) -167 N

Explanation:

Given: m = 0.530 kg; vi = 18.0 m/s; vf = 13.5 m/s; t = 0.100 s

Find: (a) Impulse, (b) Force

(a) Impulse = Momentum Change = m•Delta v = m•(vf - vi)= (0.530 kg)•( -13.5 m/s - 18.0 m/s)

Impulse = -16.7 kg•m/s = -16.7 N•s

where the "-" indicates that the impulse was opposite the original direction of motion.

(Note that a kg•m/s is equivalent to a N•s)

(b) The impulse is the product of force and time. So if impulse is known and time is known, force can be easily determined.

Impulse = F•t

F = Impulse/t = (-16.7 N s) / (0.100 s) = -167 N

where the "-" indicates that the impulse was opposite the original direction of motion.

ill do it for 100, also this isnt 80 its 40

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