Conservation Of Kinetic Energy Problem

[FredericChopin]

Homework Statement

A body of mass 2.0 kg makes an elastic collision with another body at rest and continues to move in the same direction but with one-fourth of its original speed. What is the mass of the struck body?

Homework Equations

Conservation of Kinetic Energy

E_K = (1/2)*m*v^2

(1/2)*m₁*u₁^2 + (1/2)*m₂*u₂^2 = (1/2)*m₁*v₁^2 + (1/2)*m₂*v₂^2

The Attempt at a Solution

m₁ is given to be 2 kg, u₂ is given to be 0 m/s, v₁ is given to be 1/4 of u₁ and because of this, v₂ must be 3/4 of u₁ (Conservation Of Kinetic Energy)(1/2)*m₁*u₁^2 = (1/2)*m₁*((1/4)*u₁)^2 + (1/2)*m₂*((3/4)*u₁)^2

(Simplifying)

(1/2)*m₁*u₁^2 = (1/2)*m₁*(u₁^2/16) + (1/2)*m₂*(9*u₁^2)/16)

(Simplifying)

(1/2)*m₁*u₁^2 = (1/2)*m₁*(u₁^2/16) + (1/2)*m₂*(9*u₁^2)/16)

(Simplifying)

(1/2)*m₁*u₁^2 = (m₁*u₁^2)/32 + (9*m₂*u₁^2)/32

(Changing 1/2 to 16/32)

(16*m₁*u₁^2)/32 = (m₁*u₁^2)/32 + (9*m₂*u₁^2)/32

(Subtracting (m₁*u₁^2)/32 from both sides of the equation)

(15*m₁*u₁^2)/32 = (9*m₂*u₁^2)/32

(Multiplying both sides of the equation by 32)

15*m₁*u₁ = 9*m₂*u₁^2

(Substituting m₁ = 2 kg)

30*m₁*u₁ = 9*m₂*u₁^2

(Dividing both sides of the equation by 9*u₁^2)

m₂= 3.33 kgUnfortunately, the answer was 1.2 kg. I think the problem was that I falsely assumed that if the first body had a final velocity one-fourth that of its initial velocity, then the second body would have three-fourths that of the first body's initial velocity, however this still makes sense since the second body was a rest before the collision and the collision was elastic.

I have been struggling to get this question right for 3 days!

Hints, anybody?

Thank you.

 

[Doc Al]

m₁ is given to be 2 kg, u₂ is given to be 0 m/s, v₁ is given to be 1/4 of u₁ and because of this, v₂ must be 3/4 of u₁ (Conservation Of Kinetic Energy)

No, that's not correct. You should solve for v₂ and see what it actually equals. (KE is conserved, not velocity.)

 

[Doc Al]

While it's true that $1 = 1/4 + 3/4$,

It is not true that $1^2 = (1/4)^2 + (3/4)^2$.

 

[CAF123]

You can use first the conservation of momentum (taking your system as the two cars) and then conservation of kinetic energy. The former will allow you to obtain an expression for the velocity of the second car after the collision.

 

[DEvens]

In 1-D collisions of this nature, it is usually easier to transform to the centre of momentum frame. In that frame, an elastic collision is very simple. Then transform back to the lab frame. So the magnitudes of the COM frame velocities before the collision are gotten from these.

Vc1 m1 = Vc2 m2 (momentums total to zero in the COM frame)
Vc1 + Vc2 = V (closing speed still V)

Vc1 is the COM speed of mass 1
Vc2 is the COM speed of mass 2

After the collision, in the COM frame, mass m1 simply flips its direction, keeping the same speed. And after the collision its speed in the lab frame is V/4. But the speed at which the COM frame is moving relative to the lab frame is V - Vc1. So:

V - 2 Vc1 = V/4

Those are three very easy to solve equations for the unknowns Vc1, Vc2, and m2. You know m1, and V can be eliminated.

The COM frame makes a lot of collision calculations much simpler.
Dan

 

[consciousness]

For elastic collisions the "coefficient of restitution" makes things a LOT easier.

 

[FredericChopin]

Thank you all.

I tried making v₂^2 the subject of the equation, but here is what I got:

(1/2)*m₁*u₁^2 = (1/2)*m₁*((1/4)*u₁)^2 + (1/2)*m₂*v₂^2

(Simplifying)

(1/2)*m₁*u₁^2 = (1/2)*m₁*(u₁)^2/16) + (1/2)*m₂*v₂^2

(Simplifying)

(1/2)*m₁*u₁^2 = (m₁*u₁^2)/32 + (1/2)*m₂*v₂^2

(Changing 1/2 to 16/32)

(16*m₁*u₁^2)/32 = (m₁*u₁^2)/32 + (16*m₂*v₂^2)/32

(Subtracting (m₁*u₁^2)/32 from both sides of the equation)

(15*m₁*u₁^2)/32 = (16*m₂*v₂^2)/32

(Multiplying both sides of the equation by 32)

15*m₁*u₁^2 = 16*m₂*v₂^2

(Substituting m₁ = 2 kg)

30*u₁^2 = 16*m₂*v₂^2

(Dividing both sides of the equation by 16*m₂)

v₂^2 = (30*u₁^2)/(16*m₂)

(Simplifying)

v₂^2 = (15*u₁^2)/(8*m₂)Hm... I'm either not looking at it right or this doesn't tell me much since I still don't know u₁. I tried substituting it back into the original equation but the m₂'s cancel each other out.

Ugh... 4 days now...

 

[Doc Al]

I tried making v₂^2 the subject of the equation, but here is what I got:

(1/2)*m₁*u₁^2 = (1/2)*m₁*((1/4)*u₁)^2 + (1/2)*m₂*v₂^2

That's fine.

Hm... I'm either not looking at it right or this doesn't tell me much since I still don't know u₁. I tried substituting it back into the original equation but the m₂'s cancel each other out.

You've used conservation of KE, but that's not enough by itself. What else is conserved in any collision?

 

[FredericChopin]

Momentum is conserved!

O.K., let's see if it works:

If v₂ = √((15*u₁²)/(8*m₂))

, then:

m₁*u₁ + m₂*u₂ = m₁*v₁ + m₂*v₂

(Substituting u₂ = 0 m/s, v₁ = (1/4)*u₁ and v₂² = √((15*u₁²)/(8*m₂)))

m₁*u₁ = m₁*(1/4)*u₁ + m₂*√((15*u₁²)/(8*m₂))

(Simplifying)

m₁*u₁ = (m₁*u₁)/4 + m₂*√((15*u₁²)/(8*m₂))

(Changing 1/1 to 4/4)

(4*m₁*u₁)/4 = (m₁*u₁)/4 + m₂*√((15*u₁²)/(8*m₂))

(Subtracting (m₁*u₁)/4 from both sides of the equation)

(3*m₁*u₁)/4 = m₂*√((15*u₁²)/(8*m₂))

(Squaring both sides of the equation)

(9*m₁²*u₁²)/16 = m₂²*(15*u₁²)/(8*m₂)

(Simplifying)

(9*m₁²*u₁²)/16 = (15*u₁²*m₂²)/(8*m₂)

(Changing 15/8 to 30/16)

(9*m₁²*u₁²)/16 = (30*u₁²*m₂²)/(16*m₂)

(Simplifying)

(9*m₁²*u₁²)/16 = (30*u₁²*m₂)/16

(Multiplying both sides of the equation by 16)

9*m₁²*u₁² = 30*u₁²*m₂

(Substituting m₁ = 2 kg)

36*u₁² = 30*u₁²*m₂

(Dividing both sides of the equation by 30*u₁²)

m₂ = 1.2 kg


Thank you so much!

 

[Doc Al]

m₂ = 1.2 kg


Thank you so much!

Excellent!

Now it's time to learn a shortcut.

For a perfectly elastic straight-line collision, the relative velocity is reversed during the collision:

u_1 - u_2 = v_2 - v_1
(This can be proven by combining conservation of energy with conservation of momentum.)

Thus: u = v₂ - u/4
Which leads to v₂ = (5/4)u

Now combine this with conservation of momentum and you'll be able to solve for the mass in one line.

All this is perfectly consistent with what you did. (It's the same thing, really.)