Inelastic collision blocks of wood

In summary, a mass of 5g moving to the right at a speed of 100m/s collides with a block of wood that is also moving to the right at 100m/s. The speed just after the collision is 5.71m/s.
  • #1
nickname
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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
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 [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
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
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:

[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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  • #5


1. Yes, your answer for the speed after the collision is correct. In an inelastic collision, the two objects stick together after impact and move with a common velocity, which is calculated using the equation you provided.

2. To solve for the angle of deflection, we can use the conservation of energy principle. The initial kinetic energy of the system is equal to the potential energy at the maximum height:

1/2(m1+m2)v^2 = mgh

Plugging in the values we know, we get:

1/2(0.005+0.100)(5.71)^2 = (0.005+0.100)gh

Solving for h, we get:

h = 0.53 m

Now, we can use the trigonometric relationship between the length of the pendulum, the height, and the angle of deflection:

tan(theta) = h/2

Substituting the value of h, we get:

tan(theta) = 0.53/2 = 0.265

Taking the inverse tangent, we get:

theta = 14.5 degrees

So, the pendulum will deflect at an angle of approximately 14.5 degrees. Your approach of using cosine to find the angle is also correct, but it would be more complicated in this case as you would need to use the law of cosines to find the angle. Using tangent is a simpler approach in this case.
 

Related to Inelastic collision blocks of wood

What is an inelastic collision?

An inelastic collision is a type of collision in which kinetic energy is not conserved. This means that the total kinetic energy of the objects before and after the collision is not the same. Some of the kinetic energy is lost in the form of heat or sound.

How do inelastic collisions differ from elastic collisions?

In elastic collisions, kinetic energy is conserved and the objects bounce off each other without any loss of energy. In inelastic collisions, the objects stick together or deform upon impact, resulting in a loss of kinetic energy.

What is the formula for calculating the coefficient of restitution in an inelastic collision?

The coefficient of restitution (COR) is the ratio of the relative velocities of the objects before and after the collision. In an inelastic collision, the COR is less than 1, indicating a loss of kinetic energy. The formula for calculating COR is COR = (v2 - v1) / (u1 - u2), where v1 and v2 are the final velocities and u1 and u2 are the initial velocities of the objects.

How can the mass and velocity of the objects affect the outcome of an inelastic collision?

The mass and velocity of the objects have a direct impact on the amount of kinetic energy lost in an inelastic collision. Objects with larger masses and higher velocities will have a greater loss of kinetic energy compared to objects with smaller masses and lower velocities.

What are some real-life examples of inelastic collisions involving blocks of wood?

Some real-life examples of inelastic collisions involving blocks of wood include a hammer hitting a nail, a car hitting a tree, or a bowling ball hitting a set of pins. In each of these cases, the objects involved stick together or deform upon impact, resulting in a loss of kinetic energy.

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