# Rolling cart

1. Oct 18, 2004

### ElectricMile

A 1000 kg cart is rolling to the right at 1.70m/s . A 70.0 kg man is standing on the right end of the cart.What is the speed of the cart if the man suddenly starts running to the left with a speed of 7.00 m/s relative to the cart?

really boggles my mind, should i be using

=(m1 +m2)Vf - m1(Vi)/m2
or whats going on?!

2. Oct 19, 2004

### ponjavic

(m1+m2)vb=m1*v1a+m2*v2a
we put the positive direction as right
(m1+m2)vb/m1-m2*v2a/m1=v1a

Im not sure how you define the masses but i think you missed (m1 +m2)Vf/!!m2!!

3. Oct 19, 2004

### HallsofIvy

Staff Emeritus
Conservation of momentum. Initially, the 1000kg cart is rolling to the rightare 1.7 m/s while the man is standing still, so the total momentum is (mass)(velocity) 1700 kgm/s.

Let v be the speed of the cart after the man starts running. When the man (mass 70.0 kg) runs to the right at 7.00 m/s, relative to the cart, his speed relative to the ground is v- 7 and so his momentum, relative to the ground, is 70(v- 7)= 70v- 490 kgm/s. The momentum of the cart is 1000v and so the total momentum is now
1070v- 490 . By conservation of momentum, that must be
1070v- 490= 1700=> 1070v= 2190 so v= 2190/1070= 2.05 m/s to the right.

4. Oct 20, 2004

### regor60

Isn't the initial momentum a result of the mass of both the cart and the man, 1070 x1.7 or 1819 to the right ? If so, leads to 2.15 m/s

5. Oct 20, 2004

### Staff: Mentor

Yes, the initial momentum must include the man moving with the cart.