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lutik1710 [3]
3 years ago
8

a ball (mas .40) is initially moving to the left at 30m/s. After hitting the wall, the ball is moving to the right at 20 m/s. Wh

at is the impulse of the net force on the ball during its colllision with the wall
Physics
1 answer:
Step2247 [10]3 years ago
6 0

Answer:

The impulse is -20kg m/s.

Explanation:

given mass of ball =  0.4 kg

The ball is moving towards left with the initial velocity = 30m/s

The momentum in left direction P1 = mv1

= (0.4)(30)

= 12 kg m/s

The given velocity towards right = -20 m/s ( negative sign shows inverse direction)

Momentum towards right, P2 = mv2

P2 = (0.4)(-20)

P2 = -8 kg m/s

The Impulse = change in momentum

= P2 –P1

= -8-12

= -20 kg m/s

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Careful measurements have been made of Olympic sprinters in the 100-meter dash. A quite realistic model is that the sprinter's v
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Answer:

a.

\displaystyle a(0 )=8.133\ m/s^2

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c. t=9.9 \ sec

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Modeling With Functions

Careful measurements have produced a model of one sprinter's velocity at a given t, and it's is given by

\displaystyle V(t)=a(1-e^{bt})

For Carl Lewis's run at the 1987 World Championships, the values of a and b are

\displaystyle a=11.81\ ,\ b=-0.6887

Please note we changed the value of b to negative to make the model have sense. Thus, the equation for the velocity is

\displaystyle V(t)=11.81(1-e^{-0.6887t})

a. What was Lewis's acceleration at t = 0 s, 2.00 s, and 4.00 s?

To compute the accelerations, we must find the function for a as the derivative of v

\displaystyle a(t)=\frac{dv}{dt}=11.81(0.6887\ e^{0.6887t})

\displaystyle a(t)=8.133547\ e^{-0.6887t}

For t=0

\displaystyle a(0)=8.133547\ e^o

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For t=2

\displaystyle a(2)=8.133547\ e^{-0.6887\times 2}

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\displaystyle a(4)=8.133547\ e^{-0.6887\times 4}

\displaystyle a(4)=0.52\ m/s^2

b. Find an expression for the distance traveled at time t.

The distance is the integral of the velocity, thus

\displaystyle X(t)=\int v(t)dt \int 11.81(1-e^{-0.6887t})dt=11.81(t+\frac{e^{-0.6887t}}{0.6887})+C

\displaystyle X(t)=11.81(t+1.45201\ e^{-0.6887t})+C

To find the value of C, we set X(0)=0, the sprinter starts from the origin of coordinates

\displaystyle x(0)=0=>11.81\times1.45201+C=0

Solving for C

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\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15

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The equation for the distance cannot be solved by algebraic procedures, but we can use approximations until we find a close value.

We are required to find the time at which the distance is 100 m, thus

\displaystyle X(t)=100=>11.81(t+1.45\ e^{-0.6887t})-17.15=100

Rearranging

\displaystyle t+1.45\ e^{-0.6887t}=9.92

We define an auxiliary function f(t) to help us find the value of t.

\displaystyle f(t)=t+1.45\ e^{-0.687t}-9.92

Let's try for t=9 sec

\displaystyle f(9)=9+1.45\ e^{-0.687\times 9}-9.92=-0.92

Now with t=9.9 sec

\displaystyle f(9.9)=9.9+1.45\ e^{-0.687\times 9.9}-9.92=-0.0184

That was a real close guess. One more to be sure for t=10 sec

\displaystyle f(10)=10+1.45\ e^{-0.687\times 10}-9.92=0.081

The change of sign tells us we are close enough to the solution. We choose the time that produces a smaller magnitude for f(t).  

At t\approx 9.9\ sec, \text{ Lewis sprinted 100 m}

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