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qaws [65]
3 years ago
11

A 1200-kg ore cart is rolling at 10.8 m/s across a flat friction-free surface. a crane suddenly drops of ore vertically into the

cart. how fast does the cart move just after being loaded with the ore?
Physics
1 answer:
Andre45 [30]3 years ago
5 0
The value of the final speed depends on the mass of the ore.

Let's call m the mass of the ore. We can solve the exercise by requiring the conservation of momentum, which must be the same before and after the ore is loaded.

Initially, there is only the cart, so the momentum is
p=Mv=(1200 kg)(10.8 m/s)=12960 kg m/s
After the ore is loaded, the new mass will be (1200 kg+m), and the new speed is v_f. The momentum p is conserved, so it is still 12960 kg m/s. Therefore, we have
p=12960 kg m/s =(1200 kg+m)v_f
and so the final speed is
v_f =  \frac{12960 kg m/s}{1200 kg +m}
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A circuit that has little or no resistance

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3 years ago
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!

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