Block Collision: Find Distance for Elastic & Inelastic Collisions

In summary: You'll have to find a clever way to use the fact that m2 = 5m1 to find the total momentum of m1 and m2.)
  • #1
DavidAp
44
0
Block 1 of mass m1 slides from rest along a frictionless ramp from height h = 3.1 m and then collides with stationary block 2, which has mass m2 = 5m1. After the collision, block 2 slides into a region where the coefficient of kinetic friction µk is 0.35 and comes to a stop in distance d within that region. What is the value of distance d if the collision is (a) elastic and (b) completely inelastic?

Answers
(a) 0.984 m
(b) 0.246 m


Relevant Equations
m1u1 + m2u2 = m1v1 + m2v2
vf^2 = vi^2 + 2as (where s is the Δx)
_______________________________________________

This is what I attempted.
I started the problem by working on part (b) first because it seemed that finding the distance traveled through and inelastic collision would be easier due to the shared final velocity. I know that,
h = 3.1m
m2 = 5m1
μk = 0.35.

so I use the formula m1u1 + m2u2 = (m1 + m2)v. However, since block two has no initial velocity the equation becomes m1u1 = (m1 + m2)v. Replacing m1u1 with mgh I get,
v(m1 + m2) = mgh
v = m1gh/(m1 + m2)
v = m1(9.8)(3.1)/(5m1+m1)
v = 30.38m1/6m1
v = 5.06

From there I use the kinematic equation
vf^2 = vi^2 + 2as
knowing vf = 0 and, by searching the web I found that F = ma = -μkmg making, a = -μkg I plug in my knowns into the equation.
s = -vi^2/-μkg
s = -(5.06^2)/-0.35(9.8)
s = 7.46

Now thank goodness I took geometry because 7.46 = 0.246 does not pass the reflexive property! A joke of course but this brings me to my question: what did I do wrong? Obviously something horrible which will dishearten physics lovers everywhere but please, show pity. I'm having difficulty understanding this.

Can somebody explain to me what I did wrong and what I should have done to get the right answer instead and why (especially why! Baby words please!). I will very much appreciate your help. Thank you for taking the time to review my question.
 
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  • #2


DavidAp said:
This is what I attempted.
I started the problem by working on part (b) first because it seemed that finding the distance traveled through and inelastic collision would be easier due to the shared final velocity. I know that,
h = 3.1m
m2 = 5m1
μk = 0.35.

so I use the formula m1u1 + m2u2 = (m1 + m2)v. However, since block two has no initial velocity the equation becomes m1u1 = (m1 + m2)v. Replacing m1u1 with mgh I get,
There's your problem. m1u1 is a momentum; mgh is an energy. You cannot set one equal to another.

Use mgh to find m1's speed at the bottom, then you can calculate m1u1.
 

1. What is a block collision?

A block collision is a type of collision that occurs between two blocks or objects. It involves the transfer of energy and momentum between the blocks, resulting in a change in their velocities.

2. What is the difference between elastic and inelastic collisions?

In an elastic collision, the two objects bounce off each other with no energy loss. In an inelastic collision, some of the kinetic energy is lost as heat or sound, and the objects stick together after the collision.

3. How is distance calculated for elastic and inelastic collisions?

For elastic collisions, the distance can be calculated using the equations of conservation of momentum and energy. For inelastic collisions, the distance can be calculated using the coefficient of restitution, which is a measure of how much energy is lost during the collision.

4. What factors affect the distance in a block collision?

The distance in a block collision can be affected by the mass, velocity, and angle of the objects involved, as well as the type of collision (elastic or inelastic) and any external forces acting on the objects.

5. Why is it important to understand block collisions?

Understanding block collisions is important in many fields of science, such as physics, engineering, and materials science. It helps us understand how energy and momentum are transferred between objects, and how different materials behave under different collision scenarios.

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