# physics

A man of mass 70 kg and a boy of mass 30 kg dive of the end of a boat, of mass 150 kg with a horizontal velocity of 2 m/s relativeto the boat. if initially the boat is at rest, findits velocity just after (1) both the man and boy dive off simultaneously (2) the man dives of first followed by the boy

1. ๐ 0
2. ๐ 0
3. ๐ 41
1. Conserve momentum. (Initial = final)

(1) Man and boy simultaneously jump and push the boat back.
m1 = mass of man
m2 = mass of boy
v1 = velocity of man (and boy)
m3 = mass of boat
v3 =
0 = (m1+m2) v1 + m3 v3
Rearrange:
v3 = - (m1+m2) v1 / m3
= - 100 * 2 / 150
= 1.33 m/s

(2) First, the man pushes back the boat and the boy, then the boy further pushes back the boat.

Key point: When the boy jumps off at v1 _relative_ to the boat, the boat is not at rest.

m1 = mass of man
m2 = mass of boy
v1 = velocity of man
v2 = intermediate velocity of boat
m3 = mass of boat
v4 = final velocity of boat

First jump:
0 = m1 v1 + (m2+m3) v2
Second jump:
(m2+m3) v2 = m2 (v2+v1) + m3 v4

Rearrange:
v2 = - m1 v1 / (m2+m3)
v4 = v2 - (m2 v1)/ m3

Substitute first into the second
v4 = - (m1/(m2+m3)+m2/m3) v1

1. ๐ 0
2. ๐ 0
posted by Graham

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