How Is Momentum Conserved in a One-Dimensional Collision?

  • Thread starter Thread starter Soph
  • Start date Start date
  • Tags Tags
    Collision
AI Thread Summary
In a one-dimensional collision involving two blocks, the total momentum before the collision can be calculated using the equation p = mv. The initial momentum for the system is determined to be -93.7 kg·m/s, based on the masses and velocities of both blocks. To find the final velocities after the collision, the equations for initial momentum and kinetic energy must be solved simultaneously. The discussion highlights the challenge of substituting values from these equations to derive the final velocities. It concludes that, assuming no heat loss, the momentum of the system is conserved after the collision.
Soph
Messages
4
Reaction score
0

Homework Statement


A block, which has a mass of m1 = 9.50 kg is moving towards the left with a velocity of v1 = 7.20 m/s. When it collides with a second block, which has a mass of m2 = 5.50 kg, it is also moving towards the left, but with a velocity of v2 = 4.60 m/s.

What will be the total momentum of this system after the blocks collide?

Homework Equations


p = mv
Ek = 1/2 mv^2

The Attempt at a Solution



What I did was write out the equation of the initial momentum that equals to the final:

(9.5)(-7.2) + (5.5)(-4.6) = (9.5)v1prime + (5.5)v2prime

giving me: -93.7 = (9.5)v1prime + (5.5)v2prime

I also wrote out the initial and final kinetic energy equation:

(0.5)(9.5)(-7.2)^2 + (0.5)(5.5)(-4.6)^2 = (0.5)(9.5)v1prime^2 + (0.5)(5.5)v2prime^2

giving me: 304.43 = 4.75v1prime^2 + 2.76prime^2

I'm stuck on the process of substituting the two equations to find the final velocities of the two objects. Could someone lead me through the process of substituting the numbers or is there another way of looking at this?
 
Physics news on Phys.org
What is the momentum of the system before the collision? Assuming no heat loss during the collision, what can we conclude about the momentum of the system after the collision?
 
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

Momentum: why don't the two carts become one?
Replies
32
Views
1K
A bullet collides perfectly elastically with one end of a rod
Replies
21
Views
2K
How Does Angular Momentum Conservation Affect Asteroid Collision Dynamics?
7
Replies
335
Views
14K
Ambiguous question on momentum and how it works...
Replies
13
Views
1K
1-D Perfectly Elastic Collison PLEASE CHECK THANKS
Replies
3
Views
3K
Potential Energy - 2D inelastic collision - ballistic pendulum
Replies
10
Views
2K
Elastic collision of block on block
Replies
4
Views
2K
How Do You Solve for Final Velocities in Elastic Collisions?
Replies
20
Views
2K
Momentum and Elastic Collisions
Replies
6
Views
3K
Complex collision with masses and Velcro
Replies
4
Views
2K

Hot Threads

Recent Insights

Back
Top