Question

After a 70 kg person steps from a boat onto the shore, the boat moves away with a speed of 0.7 m/s with respect to the shore. If the same person steps from the shore onto the boat, the person and the boat move away from the shore with a speed of 0.5 m/s. Assume that the initial speed of the person with respect to the shore is the same in both cases and that the person is moving in a horizontal direction. (a) Draw impulse-momentum bar charts for both situations. (b) Determine the mass of the boat and the speed of the person.

Answer 1

(a) The impulse - momentum bar chart for the two situations is in the image attached.

(b) The mass of the boat is 175 kg and the speed of the person is 1.75 m/s.

How to calculate the mass of the boat and speed of the person?

The mass of the boat and speed of the person is calculated by applying the principle of conservation of linear momentum.

let the mass of the boat = m

let the speed of the person = v

first situation

(70 kg)v = (0.7 m/s ) x m

70v = 0.7m -------- (1)

second situation

(70 kg + m)0.5 m/s = (70 kg) v

(70 + m)0.5 = 70v ----------- (2)

substitute (1) into (2);

(70 + m) 0.5 = 0.7 m

35 + 0.5m = 0.7m

35 = 0.2 m

m = 35/0.2

m = 175 kg

The speed of the person;

70v = 0.7(175)

v = (0.7 x 175 ) / 70

v = 1.75 m/s

The impulse - momentum for 1st situation;

P = 70v

P = 70 kg x 1.75 m/s

P = 122.5 kgm/s

The impulse - momentum for 2nd situation;

P = (70 + m)0.5

P = (70 kg + 175 kg) 0.5 m/s

P = 122.5 kgm/s