Calculating Mass and Velocity in Elastic Collisions: Lab Cart A and B Example

[runningirl]

Homework Statement

Lab cart A(v=1.5 m/s; m=.25 kg) collides elastically with lab cart B (v=0m/s; m=?). After the collision, the velocity of cart A is -.67 m/s. Cart B then goes on to have a sticky collision with cart C (v=0 m/s; m=.50 kg). After the sticky collision, the velocity of the combined carts B and C is 0.47 m/s. What is the mass of cart B? What as the velocity of cart B right after cart A collided with it?




3. The Attempt at a Solution

m1v1+m2v2=m1vf1+m2vf2

.25(1.5)+m(0)=.25(-.67)+m(vf2)

mv1+.5(0)=(m+.5)(.47)

i wanted to substitute m or v to solve for them each...

i got m=.65 kg and v=.83 m/s.

but i wasn't sure if my method was correct.

 

[issacnewton]

Lab cart A(v=1.5 m/s; m=.25 kg) collides elastically with lab cart B (v=0m/s; m=?). After the collision, the velocity of cart A is -.67 m/s. Cart B then goes on to have a sticky collision with cart C (v=0 m/s; m=.50 kg). After the sticky collision, the velocity of the combined carts B and C is 0.47 m/s. What is the mass of cart B? What as the velocity of cart B right after cart A collided with it?

m1v1+m2v2=m1vf1+m2vf2

.25(1.5)+m(0)=.25(-.67)+m(vf2)

mv1+.5(0)=(m+.5)(.47)

i wanted to substitute m or v to solve for them each...

i got m=.65 kg and v=.83 m/s.
.

you see, the velocity of the cart B after it collides with cart A (vf2) is same as the velocity of the cart B before it collides with cart C (v1) which I have highlighted above. Since you have already used the symbol v1 as the velocity of the cart A before it collides with the cart B, don't use it again for the velocity of the cart B before it collides
with cart C.

 

[runningirl]

thanks. but is my method correct?

 

[issacnewton]

yes, so you have now two equations and two unknowns, mass of cart B (m) and the velocity of the cart B after the collision with cart A and before the collision with cart C (vf1). so solve for these.

 

[rwisz]

Your method is correct. The equation you used, m1v1 + m2v2 = m1vf1 + m2vf2, is the conservation of momentum equation for an elastic collision. This means that the total momentum before the collision is equal to the total momentum after the collision. In your case, the momentum of cart A before the collision is 0.375 kg*m/s (0.25 kg * 1.5 m/s) and after the collision it is -0.1675 kg*m/s (0.25 kg * -0.67 m/s). This change in momentum must be equal to the momentum of cart B after the collision, which is 0.47 kg*m/s (m * 0.83 m/s). Solving for m, we get m = 0.65 kg, which matches your answer. Similarly, you can use the conservation of momentum equation for the sticky collision between carts B and C to solve for the final velocity of the combined carts, which is 0.47 m/s. Great job!