A 4.0 kg ball moving to the right at 1.2 m/s collides inelastically with a 1.0 kg ball at rest. What will the velocity of the combined balls be after the collision?

Question

kittiwitti1 · Answer

Let's list out the values that we've been given.$$m_A=4.0~kg$$$$v_A=1.2~m/s$$$$m_B=1.0~kg$$$$v_B=0~m/s$$$$m_T=4.0~kg+1.0~kg=5.0~kg$$*****
$$Inelastic~collision~formula:~m_Av_A+m_Bv_B=m_Tv_T$$Plug the given values into the equation:$$(4.0\times 1.2)+(1.0\times 0)=5.0v_T$$You should be able to solve for the variable now:$$4.8=5.0v_T$$

kenljw · Answer

I know your using conservation of momentum, but?
How do you know if all the momentum is not transfer to mB?
or split between the two instead of both with same final velocity?

anonymous · Answer

Wholly inelastic collision means the balls stick together, thus they have a joint mass and one joint velocity.

kittiwitti1 · Answer

Well, from what I've learned (most recently - we've just taken the test) about momentum and collisions, inelastic collisions all follow that formula.
@douglaswinslowcooper

anonymous · Answer

Some collisions could be partially elastic or partially inelastic, with energy lost in the impact but the objects separating afterward.

kittiwitti1 · Answer

Well, yes; however we haven't learned anything of that sort.

kittiwitti1 · Answer

& it was the same kind problem

anonymous · Answer

The question is rather artificial, as balls stuck together would no longer roll.

kittiwitti1 · Answer

Really? then my textbook's ideas are all wrong?

anonymous · Answer

What would two balls stuck together look like? Unless they stuck together side by side and rolled at right angles to their axis, they would not role at all. 

Furthermore, when you get to energy conservation, in reality you have to take into account energy of rotation, too. Oh, well. The textbook writers know this. They just don't want to complicate things.