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a small 100g cart is movine at 1.20 m/s on an air track when it collides with a larger, 1.00 kg cart at rest. After the collision, the small cart recoils at 0.850 m/s. what is the speed of the large cart after the collision?

physics
Guest Feb 27, 2015

Best Answer 

 #1
avatar+808 
+5

If you haven't already I'd highly recommend making a drawing. 

 

i = initial

f = final

 

m(small cart) = 100 g = 0.1 Kg

vi(small cart) = 1.20 m/s

vf(small cart) = - 0.850 m/s

 

m(large cart) = 1.00 Kg 

vi(large cart) = 0 m/s

vf(large cart) = ?

 

pi(small cart) = mv = 0.120 Kgm/s

pi(large cart) = mv = 0 Kgm/s

pi(large cart + small cart) = 0.12 Kgm/s = pf(large cart + small cart)

pf(small cart) = mv = - 0.0850 Kgm/s 

pf(large cart) = pf(large cart + small cart) - pf(small cart) = 0.12 Kgm/s - (- 0.0850 Kgm/s) = 0.205 Kgm/s

vf(large cart) = pf(large cart)/m(large cart) = 0.205 m/s

Tetration  Feb 27, 2015
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1+0 Answers

 #1
avatar+808 
+5
Best Answer

If you haven't already I'd highly recommend making a drawing. 

 

i = initial

f = final

 

m(small cart) = 100 g = 0.1 Kg

vi(small cart) = 1.20 m/s

vf(small cart) = - 0.850 m/s

 

m(large cart) = 1.00 Kg 

vi(large cart) = 0 m/s

vf(large cart) = ?

 

pi(small cart) = mv = 0.120 Kgm/s

pi(large cart) = mv = 0 Kgm/s

pi(large cart + small cart) = 0.12 Kgm/s = pf(large cart + small cart)

pf(small cart) = mv = - 0.0850 Kgm/s 

pf(large cart) = pf(large cart + small cart) - pf(small cart) = 0.12 Kgm/s - (- 0.0850 Kgm/s) = 0.205 Kgm/s

vf(large cart) = pf(large cart)/m(large cart) = 0.205 m/s

Tetration  Feb 27, 2015

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