Conservation of linear momentum at the skating rink

AI Thread Summary
In the discussion about conservation of linear momentum at the skating rink, the problem involves a girl and her father sliding together on ice, with the girl initially moving at 3.0 m/s. The combined mass of the girl and her father is used to calculate their common speed after they connect, resulting in a speed of 0.84 m/s, which matches the book's answer. The second part of the problem involves the father pushing the daughter, leading to confusion about the calculation of their new speeds. The correct approach emphasizes the conservation of momentum, and the solution involves setting up the equation to express their velocities after the push. Ultimately, the answer key indicates their speeds should be approximately 1.9 m/s after the push.
mstud
Messages
69
Reaction score
0

Homework Statement



a) At the skating rink, a girl of 35 kgs slides at a velocity of 3.0 m/s against her father. Her father, who is standing with his skates in the direction of motion, has a mass of 90 kgs. He takes hold of her and they slide together across the ice. Friction is ignored. What is their common speed?

b) While they're gliding along at this speed, the father shoves his daughter, so that they move in opposite directions. The father still moves with the same velocity as his daughter, but in the other direction. What is their speed now.



Homework Equations



p=p0

The Attempt at a Solution



a) I have calculated their speed while moving as one body : p=p_0\Leftrightarrow (m+M)v=mv_0+0\Leftrightarrow v=(mv_0)/(m+M)=\frac{35 kg\cdot 3.0 m/s}{35 kg+90 kg}=0.84 m/s. This complies with the answer key of my book.

b) I don't manage to calculate the right answer here, the answer key of my book says it shall be approximately 1.9 m/s. Could you give me some directions?
 
Physics news on Phys.org
mstud said:
b) I don't manage to calculate the right answer here, the answer key of my book says it shall be approximately 1.9 m/s. Could you give me some directions?
Momentum is still conserved. See if you can express the given conditions mathematically.
 
I see

I must solve:
(m+M)v=Mv_2 +m(-v_2) for v_2

When I posted the original question I had tried to "conserve" kinetical energy and solve it for velocity. :lol:
 
Thanks a lot!
 
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}...
Back
Top