Time remaining:
physics problem about linear momentum

label Physics
account_circle Unassigned
schedule 1 Day
account_balance_wallet $5

particle 1 of mass M movies with speed v in the +x direction and has an elastic collision with particle 2 (mass=3M) that was orginally at rest. after the collision, particle 2 is moving in the +x direction. what is its speed?

Oct 23rd, 2017


Momentum is always conserved.
Total momentum before = total momentum after
mv = m(v1-v1 after collision) + (3m)(v2-v2 after collision)
Simplify:
v =( v1-v1 after collision)+ 3(v2-v2 after collision)

In elastic collisions, KE is conserved.
Total KE before = total KE after collision
½mv² = ½m(v1-v1 after collision)² + ½(3m)(v2-v2 after collison)²
Simplify:
v² = (v1-v1 after collision)² + 3(v2- v2 after collision)²

therefore:
v =( v1-v1 after collision) + 3(v2_v2 after collision)
v² = (v1-v1 after collision)² + 3(v2_v2after collision)²

Apr 13th, 2015

Momentum is always conserved.
Total momentum before = total momentum after
mv = m(v1-v1 after collision) + (3m)(v2-v2 after collision)
Simplify:
v =( v1-v1 after collision)+ 3(v2-v2 after collision)

In elastic collisions, KE is conserved.
Total KE before = total KE after collision
½mv² = ½m(v1-v1 after collision)² + ½(3m)(v2-v2 after collison)²
Simplify:
v² = (v1-v1 after collision)² + 3(v2- v2 after collision)²

therefore:
v =( v1-v1 after collision) + 3(v2_v2 after collision)

v1=v- 3(v2-v2 after collision) +v1 after collision

v2=v/3-( v1-v1 after collision)/3+v2 after collision


v² = (v1-v1 after collision)² + 3(v2-v2after collision)²

Apr 13th, 2015

Did you know? You can earn $20 for every friend you invite to Studypool!
Click here to
Refer a Friend
...
Oct 23rd, 2017
...
Oct 23rd, 2017
Oct 24th, 2017
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer