Inelastic collisions between two blocks

[lilblevs11]

Homework Statement

A 2.0 kg block slides along a frictionless surface at 1.0 m/s. A second block, sliding at a faster 4.0 m/s, collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.0 m/s. What is the mass of the second block?

Homework Equations

law of conservation of momentum...

m1u1 + m2u2 = m1v1 + m2v2

The Attempt at a Solution

(2 kg * 1 m/s ) + (m2 * 4 m/s) = (2kg + m2) 2.0 m/s

2 J + (4 m/s m2) = 4 J + (2.0 m/s m2)

4 m/s m2 = 2 J + (2.0 m/s m2)

now I'm lost...

 

[netgypsy]

Just finish your algebra

by the way kgm/sec is not joule, it's just kg m/sec, which is the same as N sec

so you have the equivalent of

4m = 2 + 2m

subtracting 2m from both sides
2m = 2

m = 1kg

 

[lilblevs11]

Thanks netgypsy... now i feel like a true student. haha

 

[netgypsy]

Good job!

 

[asteriatic]

I would like to point out that the approach taken in the solution attempt is correct, but the units used in the calculations are incorrect. The unit for momentum is kg*m/s, not J (Joules). Therefore, the correct equation should be:

(m1u1 + m2u2) = (m1v1 + m2v2)

(2 kg * 1 m/s) + (m2 * 4 m/s) = (2 kg * 2.0 m/s) + (m2 * 2.0 m/s)

2 kg*m/s + 4 m2/s = 4 kg*m/s + 2 m2/s

2 m2/s = 2 kg*m/s

m2 = 1 kg

Therefore, the mass of the second block is 1 kg. This means that the second block is half the mass of the first block. This result is consistent with the fact that the final velocity of the combined blocks is closer to the initial velocity of the first block (1 m/s) than the initial velocity of the second block (4 m/s). This indicates that the first block has a greater influence on the final velocity, which is expected since it has a greater mass.