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GuDViN [60]
2 years ago
6

In a common but dangerous prank, a chair is pulled away as a person is moving downward to sit on it, causing the victim to land

hard on the floor. Suppose the victim falls by 0.50 m, the mass that moves downward is 79.0 kg, and the collision on the floor lasts 0.0750 s. What are the magnitudes of the (a) impulse and (b) average force acting on the victim from the floor during the collision?
Physics
1 answer:
Travka [436]2 years ago
5 0

Answer:

A)247.4kgm/s b) 3299.0N

Explanation:

The potential energy of the victim = mgh where m is mass, g is acceleration due to gravity and h is height from where the victim fell

Pe = 79*9.81*0.5 = 1/2mv2 (kinetic energy of the motion) v is speed in m/s

v = √9.81 = 3.132m/s

Change in momentum of the victim = m(vi-vf)

Vi is zero and vf is the velocity of impact

Change in momentum = 79*(3.132-0) = 247kgm/s which also the impulse

B) average force x time = impulse = 247

Average force = 247/0.075 = 3299.04N

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You are coasting on your 12-kg bicycle at 13 m/s and a 5.0-g bug splatters on your helmet. The bug was initially moving at 1.5 m
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Answer:

a) Pi,c = 1066 kgm/s

b) Pi,b = 0.0075 kgm/s  

c) ΔV = - 0.0007 m/s

d) ΔV = - 0.0008 m/s

Explanation:

Given:-

- The mass of the bicycle, mc = 12 kg

- The mass of passenger, mp = 70 kg

- The mass of the bug, mb = 5.0 g

- The initial speed of the bicycle, vpi = 13 m/s

- The initial speed of the bug, vbi = 1.5 m/s

Find:-

a.What is the initial momentum of you plus your bicycle?

b.What is the initial momentum of the bug?

c.What is your change in velocity due to the collision the bug?

d.What would the change in velocity have been if the bug were traveling in the opposite direction?

Solution:-

- First we will set our one dimensional coordinate system, taking right to be positive in the direction of bicycle.

- The initial linear momentum (Pi,c) of the passenger and the bicycle would be:

                       Pi,c = vpi* ( mc + mp)

                       Pi,c = 13* ( 12+ 70 )

                       Pi,c = 1066 kgm/s  

- The initial linear momentum (Pi,b) of the bug would be:

                       Pi,b = vbi*mb

                       Pi,b = 0.005*1.5

                       Pi,b = 0.0075 kgm/s  

- We will consider the bicycle, the passenger and the bug as a system in isolation on which no external unbalanced forces are acting. This validates the use of linear conservation of momentum.

- The bicycle, passenger and bug all travel in the (+x) direction after the bug splatters on the helmet.

                       Pi = Pf

                       Pi,c + Pi,b = V*(mb + mc + mp)

Where,    V : The velocity of the (bicycle, passenger and bug) after collision.

                      1066 + 0.0075 = V*( 0.005 + 12 + 70 )

                      V = 1066.0075 / 82.005

                      V = 12.9993 m/s

- The change in velocity is Δv = 13 - 12.9993 =  - 0.00070 m/s      

- If the bug travels in the opposite direction then the sign of the initial momentum of the bug changes from (+) to (-).

- We will apply the linear conservation of momentum similarly.

                      Pi = Pf

                      Pi,c + Pi,b = V*(mb + mc + mp)        

                      1066 - 0.0075 = V*( 0.005 + 12 + 70 )

                      V = 1065.9925 / 82.005

                      V = 12.99911 m/s

- The change in velocity is Δv = 13 - 12.99911 =  -0.00088 m/s      

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