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Paul [167]
2 years ago
12

A uniform thin rod of length 0.60 m and mass 3.5 kg can rotate in a horizontal plane about a vertical axis through its center. T

he rod is at rest when a 3.0 g bullet traveling in the rotation plane is fired into one end of the rod. As viewed from above, the bullet's path makes angle θ = 60° with the rod. If the bullet lodges in the rod and the angular velocity of the rod is 14 rad/s immediately after the collision, what is the bullet's speed just before impact? (Answer is in m/s)
Physics
1 answer:
abruzzese [7]2 years ago
6 0
 <span>The moment of inertia for the rod is 
mr*L^2/12 
After collision 
I=mr*l^2/12+mb*r^2 
I=0.073 
The momentum of the rod after collision is 
I*w 
=1.022 

The momentum of the bullet that is radial 
to the rod is 
vr=v*sin(60) 
vr/r=w 

so the radial momentum of the bullet is 
3*vr/(0.25*1000) 

vr*0.012 

applying conservation of momentum 
vr*0.012=1.022 
vr=85 m/s 

and 
v=85/sin(60) 
v=98 m,/s 

j</span>
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A billiard ball is dropped from a height of 64 feet. Use the position function s(t) = –16???? 2 + ????0???? + ????0 to answer th
Delicious77 [7]

Answer:

s(t) = -16*t^2 + 64

v(t) = -32*t

a(t) = -32 ft/s^2

v(t) = 64 ft/s ... At impact

Explanation:

Given:-

- The height of the billiard ball t = 0 , h = 64 ft.

- The position function of an object under gravity is given by:

                                    s(t) = -16*t^2 + v_o*t + s_o

Find:-

a. Determine the position function s(t),

b. the velocity function v(t),

c. the acceleration function a(t).

d. What is the velocity of the ball at impact?

Solution:-

- To determine the position function we must initialize our problem and use the given general equation.

- s(t) is the position of the billiard ball from the ground at time t. So when t = 0, then s(t) = h. Hence, we have:

                                  s(t) = s_o = h = 64 ft

- Similarly we know that v_o is the initial velocity of the ball. Since, the ball was dropped we say that the initial velocity v_o = 0. Hence, the position of the ball from ground is given by following expression:

                                  s(t) = -16*t^2 + 64  

- To find the velocity expression v(t) we will take the time derivative of the position expression s(t) as follows:

                                  v(t) = d s(t) / dt

                                  v(t) = -16*2*t + 0

                                  v(t) = -32*t ft/s

- Similarly, the expression for acceleration a(t) is given by the time derivative of the velocity expression v(t) as follows:

                                  a(t) = d v(t) / dt

                                  a(t) = -32*t

                                  a(t) = -32 ft/s^2

- The velocity of ball at impact can be determined by evaluating s(t) = 0 and find the value for time t. Then that time t can be substituted in the velocity expression v(t) for final velocity. Or we could use the following 3rd kinematic equation as follows:

                                 v(t)^2 - 0^2 = 2*a(t)*s_o

                                 v(t)^2 = 2*(32)*(64)

                                 v(t) = 64 ft/s

- The ball has a velocity of 64 ft/s at impact!

6 0
3 years ago
A car with a mass of 3 Kg and velocity of 40 m/s collided with a truck of a velocity of 60 m/s, if the momentum is conserved wha
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Answer:

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initial velocity of the car, u₁ = 40 m/s

initial velocity of the truck, u₂ = 60 m/s

let the mass of the truck = m₂

Apply the principle of conservation of linear momemtum;

m₁u₁ = m₂u₂

m₂ = (m₁u₁) / u₂

m₂ = (3 x 40) / (60)

m₂ = 2 kg

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Answer:

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Explanation:

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kondor19780726 [428]
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K.E = 1/2 x 0.18 kg x 11 m/S^2

K.E = 0.99

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3 years ago
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