A lumberjack (mass = 98 kg) is standing at rest on one end of a floating log (mass = 290 kg) that is also at rest. The lumberjack runs to the other end of the log, attaining a velocity of + 3.3 m/s relative to the shore, and then hops onto an identical floating log that is initially at rest. Neglect any friction and resistance between the logs and the water.

(a) What is the velocity of the first log just before the lumberjack jumps off? (Indicate the direction of the velocity by the sign of your answers.)

m/s

(b) Determine the velocity of the second log if the lumberjack comes to rest on it.

m/s

Question

Callum Harmon

Physics

15 February, 04:38

A lumberjack (mass = 98 kg) is standing at rest on one end of a floating log (mass = 290 kg) that is also at rest. The lumberjack runs to the other end of the log, attaining a velocity of + 3.3 m/s relative to the shore, and then hops onto an identical floating log that is initially at rest. Neglect any friction and resistance between the logs and the water.

(a) What is the velocity of the first log just before the lumberjack jumps off? (Indicate the direction of the velocity by the sign of your answers.)

m/s

(b) Determine the velocity of the second log if the lumberjack comes to rest on it.

m/s

+5

Answers (1)

Know the Answer?

Jairo Wall · Answer 1

a) V = - 1.11 m/s

b) V = 0.833 m/s

Explanation:

Given that

M (Mass of lumberjack) = 98 kg

Final velocity of lumberjack, V₁ = 3.3 m/s

m (Mass of log) = 290 kg

a)

There is no any external force that is why linear momentum will be conserved

Pi=Pf

initially both are at rest so Pi=0 (P = m v)

Lets take speed of log is V

0 = M V₁ + m V

V = - M V₁ / m

V = - 98 x 3.3 / 290

V = - 1.11 m/s

b)

Now again from linear momentum conservation

(m + M) V = M V₁

V = M V₁ (m+M)

By putting the values

V = 98 x 3.3 / (98 + 290)

V = 0.833 m/s