Collision - a ball and string form a pendulum

[ron360]

Collision -- a ball and string form a pendulum...

1. 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.
a) What is the speed of the block just after the collision?
b) What is the speed of the ball just after the collision?
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?
2. PEi + KEi = PEf + KEf [/b]
3. i know the answer to a) is 4.7m/s but i keep getting it wrong

 

[paisiello2]

What answer did you get and how did you go about getting it?
Are you missing a formula somewhere?

 

[ron360]

the formula is actually: [1/2m1v1i^2 + 1/2 m2v2i^2 = 1/2m1v1f^2 + 1/2m2v2f^2] for a) and b) ,for c) i used [m*a = -u*m*g] and [vf^2 = vi^2 + 2*a*d] ,i got 4.67m/s for a) 2.319m/s for b) and 4.82 m for c) , i know a) is right but i don't have the answers for b) and c) so i want to make sure i got it right,

 

[haruspex]

the formula is actually: [1/2m1v1i^2 + 1/2 m2v2i^2 = 1/2m1v1f^2 + 1/2m2v2f^2] for a) and b) ,for c) i used [m*a = -u*m*g] and [vf^2 = vi^2 + 2*a*d] ,i got 4.67m/s for a) 2.319m/s for b) and 4.82 m for c) , i know a) is right but i don't have the answers for b) and c) so i want to make sure i got it right,

In the OP you wrote you were getting a wrong answer. Now you're saying you get a right answer and two that may be right but you don't know. Which is it?
If you want your working checked, please post all your working. Use symbolic form, only plugging in numbers at the final step of each part. Don't carry numerical results from one part into the next, except at the final step.

 

[ron360]

sorry for being unclear here is my work for parts b) and c) i just need to double check i got the right answers, thanks a lot.
since i got V2f=4.67 m/s from a)

b) 1/2m1v1i^2 + 1/2 m2v2i^2 = 1/2m1v1f^2 + 1/2m2v2f^2
1/2(1)(7) + 1/2(2)(0)^2 = 1/2(1)V1f^2 + 1/2(2)(4.67)^2
v1f = 2.319

c)Sum of the horizontal forces on the block (direction of motion is +) = m*a = -friction
m*a = -u*m*g
a = -u*g
a = -0.23 * 9.81 m/s^2 = -2.26 m/s^2

motion for the distance
vf^2 = vi^2 + 2*a*d

With vf = 0, solve for d
d = -vi^2/(2*a)

Plug in numbers
d = -(4.67 m/s)^2 / (2*-2.26 m/s^2) = 4.82 m

 

[paisiello2]

Can you show how you got a)?

What about conservation of momentum?

 

[ron360]

1/2*m*(vf^2-vi^2) + m*g*(hf-hi) = 0

m's cancel, vi = 0, and hf = 0
vf = sqrt(2*g*hi)
vf = sqrt(2*9.81 m/s^2 * 2.5 m) = 7.00 m/s m1*v1i + m2*v2i = m1*v1f + m2*v2f

KE: (with the 1/2's canceled)
m1*v1i^2 + m2*v2i^2 = m1*v1f^2 + m2*v2f^2

To make equations a little simpler, m2 = 2*m1 and v2i = 0. The m1's cancel, v2i terms go to 0, and our equations become:
v1i = v1f + 2*v2f
v1i^2 = v1f^2 + 2*v2f^2

Solve the top equation for v1f and plug into the bottom equation and solve for v2f
v1f = v1i - 2*v2f
v1i^2 = (v1i - 2*v2f)^2 + 2*v2f^2
v1i^2 = v1i^2 - 4*v1i*v2f + 4*v2f^2 + 2*v2f^2
0 = -4*v1i*v2f + 6*v2f^2
0 = -2*v2f*(2*v1i - 3*v2f)
0 = 2*v1i - 3*v2f
v2f = 2/3*v1i v2f = 2/3*7.00 m/s = 4.67 m/s

 

[paisiello2]

So you said you keep getting it wrong. Where do you think you went wrong?

I didn't check all your numbers but the only thing I can see you did wrong for step b) is miss the negative sign for v1f.

 

[haruspex]

s
since i got V2f=4.67 m/s from a)

b) 1/2m1v1i^2 + 1/2 m2v2i^2 = 1/2m1v1f^2 + 1/2m2v2f^2
1/2(1)(7) + 1/2(2)(0)^2 = 1/2(1)V1f^2 + 1/2(2)(4.67)^2
v1f = 2.319

I should have said, don't plug in numerical results from earlier parts at all .
The 4.67 is correct to two decimal places, but isn't necessarily accurate enough for later calculations. The answer to b) is closer to 2.33.
Also, as paisiello2 noted, you need to be careful with the signs. Using energy gives an ambiguous result, so use momentum.

d = -(4.67 m/s)^2 / (2*-2.26 m/s^2) = 4.82 m

I get 4.83m.
You correctly ignored the information on static friction.

 

[ron360]

thanks a lot for all the responses, you are right a) is 4.66 m/s and b) i think is actually -2.32, correct me if I am wrong about b)

 

[haruspex]

thanks a lot for all the responses, you are right a) is 4.66 m/s and b) i think is actually -2.32, correct me if I am wrong about b)

With g = 9.8m/s exactly (which it isn't, of course) , I get a) 14/3 exactly, b) -7/3 exactly. So I'd go for 4.67, -2.33.