A89.3 kg man sits on the stern of a 5.8 m
long boat. the prow of the boat touches the
pier, but the boat isn’t tied. the man notices his mistake, stands up and walks to the boat’s prow, but by the time he reaches the prow, it’s moved 4.17 m away from the pier.
assuming no water resistance to the boat’s
motion, calculate the boat’s mass (not counting the man).
answer in units of kg

34.9 kg

Explanation:

Since there are no net external forces on the system, the center of gravity does not move.

Let's say that m is the mass of the man, M is the mass of the boat, and L is the length of the boat.

When the man is at the stern, the distance between the center of gravity and the pier is:

CG = (m L + M (L/2)) / (m + M)

When the man reaches the prow, the distance between the center of gravity and the pier is:

CG = (m x + M (x + L/2)) / (m + M)

Since these are equal:

(m L + M (L/2)) / (m + M) = (m x + M (x + L/2)) / (m + M)

m L + M (L/2) = m x + M (x + L/2)

m L + M (L/2) = m x + M x + M (L/2)

m L = m x + M x

m L − m x = M x

m (L − x) = M x

M = m (L − x) / x

Plugging in values:

M = 89.3 kg (5.8 m − 4.17 m) / 4.17 m

M = 34.9 kg

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s