Answer:
(a)2.7 m/s
(b) 5.52 m/s
Explanation:
The total of the system would be conserved as no external force is acting on it.
Initial momentum = final momentum
⇒(4.30 g × 943 m/s) + (730 g × 0) = (4.30 g × 484 m/s) + (730 g × v)
⇒ 730 ×v = (4054.9 - 2081.2) =1973.7
⇒v=2.7 m/s
Thus, the resulting speed of the block is 2.7 m/s.
(b) since, the momentum is conserved, the speed of the bullet-block center of mass would be constant.
Thus, the speed of the bullet-block center of mass is 5.52 m/s.
To solve this problem it is necessary to apply the concepts related to momentum, momentum and Force. Mathematically the Impulse can be described as
Where,
F= Force
t= time
At the same time the moment can be described as a function of mass and velocity, that is
Where,
m = mass
v = Velocity
From equilibrium the impulse is equal to the momentum, therefore
PART A) Since the body ends at rest, we have the final speed is zero, so the momentum would be
Therefore the magnitude of the person's impulse is 1125Kg.m/s
PART B) From the equation obtained previously we have that the Force would be:
Therefore the magnitude of the average force the airbag exerts on the person is 45000N