Inelastic collision blocks of wood

AI Thread Summary
In the discussion about an inelastic collision involving a mass and a block of wood, the initial calculation for the speed after the collision was corrected to 4.76 m/s. The second part of the problem involved determining the angle of deflection for a ballistic pendulum, where the height was calculated as 1.16 m. The angle was then found using the cosine function, resulting in an angle of approximately 65.2 degrees. The participants emphasized the importance of algebraic clarity in their calculations. Overall, the discussion focused on correcting calculations and clarifying the methods for solving the physics problems.
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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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nickname said:
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 h + R\cos\theta 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 \theta

AM
 
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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!
 
nickname said:
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:

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

\theta = \cos^{-1}(1 -\frac{v^2}{2gR})

AM
 
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