Inelastic collision blocks of wood

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1. a mass m1=5g moving to the right at a speed of 100m/s undergoes a perfectly inelastic collision with a block of wood, m2= 100g, at rest. what is the speed just after the collision?

2. if the mass and block of wood described in question #1 are parts of a ballistic pendulum of length 2m, through what angle will the pendulum deflect?


for question #1 I got:
vf= m1v1i+m2v2i / m1+m2= 5.71m/s after converting the mass to kg

for #2 for the height

I did : v^2= 2 g h

h= 1.29m then do I have to use cos (length-height)/ length to solve for the angle or can I just use tan theta= 2/1.29
How do I solve for number #2 and why? Is my answer for part #1 correct?

Thank you!
 

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  • #2
Andrew Mason
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1. a mass m1=5g moving to the right at a speed of 100m/s undergoes a perfectly inelastic collision with a block of wood, m2= 100g, at rest. what is the speed just after the collision?

2. if the mass and block of wood described in question #1 are parts of a ballistic pendulum of length 2m, through what angle will the pendulum deflect?


for question #1 I got:
vf= m1v1i+m2v2i / m1+m2= 5.71m/s after converting the mass to kg

for #2 for the height

I did : v^2= 2 g h

h= 1.29m then do I have to use cos (length-height)/ length to solve for the angle or can I just use tan theta= 2/1.29
How do I solve for number #2 and why? Is my answer for part #1 correct?
Your method for #1 is correct but there is a problem with your math. Use MKS units: vf = (.005 * v1i )/ (.005 + .100)

For #2, write out the expression for height in terms of the angle and the length of the pendulum. This is just geometry. (hint: what does [itex]h + R\cos\theta[/itex] equal?).

As you show, kinetic energy of m1 + m2 after collision is converted to gravitational potential energy at maximum deflection. So equate the kinetic energy to potential energy to find the height (h = v^2/2g as you have shown, which is correct) with h expressed in terms of R and [itex]\theta[/itex]

AM
 
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  • #3
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thank you of replying!

Yes, I did a wrong calculation for part 1, the answer is 4.76m/s

for part 2, I did (4.76m/s)^2= 2x9.8xh so h= 1.16 and then to find angle I did cos theta= length -h/ length so theta is cos^-1= 2-1.16/2= cos^-1 (0.42) theta= 65.2 degress is this correct for part 2?

Thank you so much for your help!
 
  • #4
Andrew Mason
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Yes, I did a wrong calculation for part 1, the answer is 4.76m/s
That's what I get.

for part 2, I did (4.76m/s)^2= 2x9.8xh so h= 1.16 and then to find angle I did cos theta= length -h/ length so theta is cos^-1= 2-1.16/2= cos^-1 (0.42) theta= 65.2 degress is this correct for part 2?
That looks right. You could make your answer clearer. It is a good practice to write out the solution algebraically before plugging in the numbers:

[tex]h = v^2/2g = R(1-cos\theta)[/tex]

[tex]\theta = \cos^{-1}(1 -\frac{v^2}{2gR})[/tex]

AM
 
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