Conservation of Linear momentum

[lisastar]

Homework Statement

A 500g putty ball moving horizontally at 6m/s collides with and sticks to a block lying on a friction-less horizontal surface. If 25% of the kinetic energy is lost, what is the mass of the block?

Homework Equations

initial (i) = final (f)
m1v1+m2v2 = m1v1+m2v2

The Attempt at a Solution

(m1v1i - m1v1f ) /v2f I'm confused :-( Somebody help me solve this. Also is the reduction of the kinetic energy relevant to solving this question?

 

[jbriggs444]

initial (i) = final (f)

Ummm, initial what = final what?

m1v1+m2v2 = m1v1+m2v2

The sum of the initial momentum of the putty blob and the final momentum of the block is equal to the sum of the initial momentum of the putty blob and the final momentum of the block? That's not a principle of physics. That's just the reflexive law for equality: A quantity is always equal to itself.

It would be more helpful to equate the sum of the initial momentum of the blob plus the initial momentum of the block to the sum of the final momentum of the blob plus the final momentum of the block.

Can you write down that equality?

 

[lisastar]

That's what I wrote up there P (initial) = P (final) sorry it wasn't clear.

 

[sunnnystrong]

By conservation of momentum
m₁ u = (m₁+ m₂) v
v/u = m₁ /(m₁+ m₂) EQN 1
Kinetic Energy -->
(1/2)(m₁+ m₂) v ² = (1/2) 0.25 m₁ u² EQN 2

 

[TJGilb]

Momentum must be conserved. Therefore we can say $\vec p = m_{ball} \vec v_{ball}$ must equal $\vec p = m_{ball+block} \vec v_{ball+block}$. We also know that kinetic energy in this case is given by $KE=\frac 1 2 mv^2$. With the given data we can calculate the initial kinetic energy which we can then easily use to calculate the final kinetic energy. Since we also know the momentum as well, we can then solve for mass and velocity of the block by putting one of them in terms of the other, and plugging into the other equation.

 

[haruspex]

By conservation of momentum
m₁ u = (m₁+ m₂) v
v/u = m₁ /(m₁+ m₂) EQN 1
Kinetic Energy -->
(1/2)(m₁+ m₂) v ² = (1/2) 0.25 m₁ u² EQN 2

This is getting a bit close to solving the problem for the student. Please start with hints, like "what do you know about the two final velocities?"