Momentum problem (Car and object)

[Dorney]

[SOLVED] Momentum problem (Car and object)

1. A crash test car with mass 1,000 kg moving at a constant speed of 12 m/sec collides completely inelastically with an object of mass M at time t=0. The object was initially at rest. The speed V in m/sec of the car-object system after the collision is given as a function of time t in seconds by the expression V= 8 / (1+5t).

a) Find the mass M of the object.
b)Assuming an initial position of x=0, determine an expression for the position of the car-object system after the collision as a function of time t.
c)Determine an expression for the resisting force on the car-object system after the collision as a function of time t.
d)Determine the impulse delivered to the car-object system from t=0 to t=2 sec.


2. Equations for the conservation of momentum.
m1*v1 + m2*v2 = (m1 +m2)*v'


3. a)I found the mass M to be 500 kg because at time=0 (time of collision), V= 8 m/sec. I used the equation for perfectly inelastic collisions.
b)The derivative of position function is equal to the velocity function, so I need to integrate. When I did that, there was a zero in the bottom of the fraction. Can't divide by zero.
c) ?
d) impluse = force*time. I assume that I need to know the equation in part c to find d. Once I know the function in part c, d is found by the area under the graph.

Thanks in advance! :)

 

[Dorney]

sorry!

I realized I posted my question in the wrong spot. It is now in the Intro Physics section.

 

[Gokul43201]

3. a)I found the mass M to be 500 kg because at time=0 (time of collision), V= 8 m/sec. I used the equation for perfectly inelastic collisions.

Correct.

b)The derivative of position function is equal to the velocity function, so I need to integrate. When I did that, there was a zero in the bottom of the fraction. Can't divide by zero.

Where does the zero come from? Do you know how to integrate the function 1/t?

c) ?

Newton's second law.

d) impluse = force*time. I assume that I need to know the equation in part c to find d. Once I know the function in part c, d is found by the area under the graph.

There's an alternative definition of the impulse that makes this easier to do. If you do it your way, you'll have to utilize the first fundamental theorem of calculus.

 

[Dorney]

b) I should know how to integrate it, but in fact, I do not. Can you help me out with this?
c) F=d (mv) / dt ?
d) impulse is also equal to m*delta v
so, use the equation to find v at 2 sec. (v2-v1)*M

 

[Dorney]

d) impulse = m*(delta v)

V at t= 2 is (8/11) m/sec.
V at t= 0 is 8 m/sec.

Impulse= (1,500)*(8-(8/11))
Impulse= 10,909 kg*m / sec.

 

[Gokul43201]

d) Correct!

b) The integral of 1/y is ln(y). So, you need to make a suitable substitution (like y=1+5t) to write the integral down in this form.

c) In this case, d(mv)/dt = m dv/dt. Take the derivative of your velocity function to find out what the retarding force is.

 

[Dorney]

c) F = -12,000 / (1+5t)^2

b) x(t) = ln y. Where y = 1+5t. Can I go any further?

 

[Gokul43201]

c) F = -12,000 / (1+5t)^2

You need to be more careful.

dv/dt = d/dt (8/(1+5t)) = -8*5/(1+5t)^2 (chain rule)

b) x(t) = ln y. Where y = 1+5t. Can I go any further?

That's not the way to do it. You need to learn to find integrals by the method of substitution of variables.

You want to find:

$$x(t) = \int_0^t v(t)dt = \int_0^t \frac {8dt}{1+5t}$$

Let t' = 1+5t, then t = (t'-1)/5, so dt = ...(something in terms of dt')?

Also, you need to rewrite the limits of the integral. When t=0, t'=...? And the upper limit is t, which must be rewritten as the above function of t'.

Make all of these substitutions into the above integral so it only contains t' in it. Now you can write down the value of the integral and substitute the upper and lower limiting values.

Finally, substitute back for t' in terms of t.

 

[Dorney]

c) F = -60,000 / (1+5t)^2.

Thanks, I will look over your explanation and see if I can solve for the function.

 

[Gokul43201]

c) F = -60,000 / (1+5t)^2.

Yes.