How Does Kinetic Energy Transfer in a Perfectly Elastic Collision?

AI Thread Summary
In a perfectly elastic collision, both momentum and kinetic energy are conserved. A 0.16 kg ball collides head-on with a second ball at rest, resulting in the second ball moving at half the speed of the first. The initial kinetic energy of the first ball is calculated, and the question arises about the fraction transferred to the second ball, with an answer of 108 J. The mass of the second ball is determined to be 0.48 kg. Understanding the distinction between momentum as a vector and kinetic energy as a scalar is crucial in solving such problems.
stupif
Messages
99
Reaction score
1
1.A 0.16kg ball makes an perfectly elastic head on collision with a second ball initially at rest. the second ball moves off with half the original speed of the first ball



2. what fraction of the origianl kinetic energy gets transferred to the second ball? answer=108J
i found out the mass of the second ball, 0.48kg
help me...please
thank you



The Attempt at a Solution

 
Physics news on Phys.org
stupif said:
1.A 0.16kg ball makes an perfectly elastic head on collision with a second ball initially at rest. the second ball moves off with half the original speed of the first ball



2. what fraction of the origianl kinetic energy gets transferred to the second ball? answer=108J
i found out the mass of the second ball, 0.48kg
help me...please
thank you



The Attempt at a Solution


You are aware that both momentum and kinetic energy are conserved in a "perfrectly elastic collision" aren't you. Do you also remember that momentum is a vector and Kinetic energy is scalar.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Similar threads

Ambiguous question on momentum and how it works...
Replies
13
Views
1K
Work done during a collision -- Change in Kinetic Energy & change in Momentum
Replies
1
Views
2K
How Do You Solve for Final Velocities in Elastic Collisions?
Replies
20
Views
2K
Solving the Elastric 3-Body Collision Problem
Replies
2
Views
2K
Solving Elastic Collision of Two Balls: Theory & Solutions
Replies
15
Views
3K
Confused about a conservation of energy problem
Replies
4
Views
2K
State whether the distance x for the block would be greater/less/same
Replies
25
Views
1K
Elastic collision -- Energy & Momentum
Replies
16
Views
2K
Kinetic energy and momentum in an elastic collision
Replies
22
Views
4K
Which Has More Momentum and Kinetic Energy After a Perfectly Elastic Collision?
Replies
1
Views
1K

Hot Threads

Recent Insights

Back
Top