Elastic Collision problem

[ron360]

A 1.0-kg ball is attached to the end of a 2.5-m string to form a pendulum. This pendulum is released from rest with the string horizontal. At the lowest point in its swing when it is moving horizontally, the ball collides elastically with a 2.0-kg block initially at rest on a horizontal frictionless surface.
v1i = 7 m/s and v2i=0 m/s
v1f=-2.33 m/s and v2f=4.66 m/s

c) If the coefficient of static and kinetic frictions are 0.3 and 0.23 respectively between the block and the surface, how far does the block travel before coming to rest?

[Unity+]

A 1.0-kg ball is attached to the end of a 2.5-m string to form a pendulum. This pendulum is released from rest with the string horizontal. At the lowest point in its swing when it is moving horizontally, the ball collides elastically with a 2.0-kg block initially at rest on a horizontal frictionless surface.

v1i = 7 m/s and v2i=0 m/s

v1f=-2.33 m/s and v2f=4.66 m/s

 

c) If the coefficient of static and kinetic frictions are 0.3 and 0.23 respectively between the block and the surface, how far does the block travel before coming to rest?

I am assuming that c is the last problem and changes the scenario so that there is friction.

 

Using Newton's laws, we can first calculate the kinetic friction force(when an object is moving on the surface).

 

[math]F_{k}=\mu_{k}F_{n}=(0.23)((2kg)(9.81))[/math]

[math]F_{k}=4.5126N[/math]

 

Since we now know the force of friction upon the block, we can use it to determine the acceleration in the opposite direction of the block's movement because the force of friction upon the block reflects the force acted upon the surface(this is where others may have to correct me, but I think it is correct).

 

[math]F=ma[/math]

[math]4.5126=(2kg)a[/math]

[math]a=2.2563\frac{m}{s^{2}}[/math]

 

Assuming that the v2f represents the velocity of the block when it is hit, we can use the kinematic equations to determine the stopping point.

 

[math]v_{f}^{2} = v_{i}^{2} + 2ad[/math]

[math]0 = (4.66)^{2} + 2(2.2563)d[/math]

 

Solve for d(distance).

 

[math](4.66)^{2} = 2(2.2563)d[/math]

[math]21.7156 = 2(2.2563)d[/math]

[math]21.7156=4.5126d[/math]

[math]21.7156=4.5126d[/math]

[math]4.8122m[/math]

 

Therefore, the block should stop 4.8122m from its original position. Since the block would already be in motion, only the kinetic friction applies.

 

EDIT: Had to fix an error.

 

EDIT2: Of course, you must change the velocity of the block for the equation in order to account for the static friction force. Did you retrieve the final velocity of the block or was it a given? If retrieved, then use the same force acting upon the block by the ball except subtract the force acted by the ball by the static friction friction force to get the total force. Then, use that force to retrieve the velocity of the block. Then, replace the 4.66 velocity with the new one in the steps above.

Edited April 8, 2014 by Unity+

[ron360]

Perfect ,thank you so much i appreciate it, it is the right answer.

[Unity+]

Perfect ,thank you so much i appreciate it, it is the right answer.

I edited my post. I apologize for the misunderstanding.

 

EDIT: Was it the right answer? I thought static friction had to be accounted for the final velocity of the block.

Edited April 8, 2014 by Unity+

[ron360]

no someone else also told me i should ignore static friction

[swansont]

!

Moderator Note

Unity+, please be aware that the point of HW help is not to work a problem out for someone, it's to help them work it out for themselves.

[studiot]

I'm sorry I don't follow this.

 

 

v1i = 7 m/s and v2i=0 m/s
v1f=-2.33 m/s and v2f=4.66 m/s

 

These are these are all calculable (which means I get these answers) from the information given.

 

 

I would say a calculation scheme would run like this, parts (a) and (b) being to calculate the velocity of the ball at impact and the velocity of the ball and block immediately after impact.

 

 

1) Calculate the velocity of the ball at impact by equating the potential energy gained from its fall from its release position to the kinetic energy it has at impact. This does indeed yield 7m/s.

 

2) We may then obtain the post impact velocities of both ball and block from conservation of both energy and momentum simultaneously. This yields -2.33m/s for the rebounding ball and 4.66m/s for the block. Notice the negative sign on the velocity of the ball.

 

3) The force of static friction does no work. So no kinetic energy is lost overcoming static friction.

This is why Ron's answer is correct. The KE acquired by the block from the ball is given by using the immediate post impact velocity of the block in the usual expression for KE.

 

4) The block moves a distance d against the constant resistance, R, of kinetic friction. This therefore does do work, dissipating the KE of the block in distance d, therefore doing Rd joules of work. Equating this to the KE of the block plus should yield d.

Edited April 8, 2014 by studiot

[Unity+]

!

Moderator Note

Unity+, please be aware that the point of HW help is not to work a problem out for someone, it's to help them work it out for themselves.

I apologize. I tried to help so he could understand the content, but I accidentally went overboard.