# 35-kg child rides a relatively massless sled down a ?

A 35-kg child rides a relatively massless sled down a hill and then coasts along the flat section at the bottom, where a second 35-kg child jumps on the sled as it passes by her. If the speed of the sled is 3.5 m/s before the second child jumps on, what is its speed after she jumps on?

### 7 Answers

- Anonymous8 months ago
1.75 hkhjhkk dgdgf

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- oubaasLv 78 months ago
Momentum is conserved, therefore :

V = m*Vi / 2m = Vi/2 = 3.5/2 = 1.75 m/sec

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- 8 months ago
momentum is conserved, so

m1v1 = (m1+m2)v2

since the mass doubles, the velocity halves

v2 = 1.75m/s

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- Andrew SmithLv 78 months ago
1.75 m/s ( ie half of the velocity just before she got on )

MOMENTUM is conserved 35 * 3.5 = 70 * v -> v = 3.5 * 35/70 = 3.5/2 = 1.75 m/s

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- Anonymous8 months ago
< a relatively massless >

Something is, or isn't, massless. There's no such thing as 'relatively massless', just like there's no such thing as being slightly pregnant.

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- electron1Lv 78 months ago
Momentum is always conserved.

Initial momentum = 35 * 3.5 = 122.5 kg * m/s

Final momentum = 70 * v

v = 122.5 ÷ 70 = 17.5 m/s

The velocity of both cars after the collision is one half of velocity because the mass of both cars is twice the car of the first car. I hope this is helpful for you!

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- 8 months ago
m1 * v1 + m2 * v2 = m3 * v3

m3 = m1 + m2

v1 = 3.5

v2 = 0

m1 = 35

m2 = 35

35 * 3.5 + 35 * 0 = (35 + 35) * v

35 * 3.5 = 2 * 35 * v

3.5 = 2 * v

1.75 = v

1.75 m/s

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