A 83-kg fisherman in a 118-kg boat throws a package of mass m = 15 kg horizontal

Question

A 83-kg fisherman in a 118-kg boat throws a package of mass m = 15 kg horizontally toward the right with a speed of v_1 = 4.5m/s as in the figure below. Neglecting water resistance, and assuming the boat is at rest before the package is thrown, find the velocity of the boat after the package is thrown. An archer shoots an arrow toward a 300-g target that is sliding in her direction at a speed of 2.45 m/s on a smooth, slippery surface. The 22.5-g arrow is shot with a speed of 42.0 m/s and passes through the target, which is stopped by the impact. What is the speed of the arrow after passing through the target? A railroad car of mass 2.50 times 10^4 kg moving at 3.35 m/s collides and couples with two coupled railroad cars, each of the same mass as the single car and moving in the same direction at 1.20 m/s. (a) What is the speed of the three coupled cars after the collision? (b) How much kinetic energy is lost in the collision?

Explanation / Answer

7.

M1 = mass of boat = 118 kg

M2 = mass of fisherman = 83 kg

m = mass of package = 15 kg

Vi = initial velocity of the combination = 0

V1 = velocity of package towards right = 4.5 m/s

V' = velocity of boat + fisherman after package was thrown

using conservation of momentum

(M1 + M2 + m) Vi = (M1 + M2) V' + m V1

(118 + 83 + 15) (0) = (118 + 83) V' + 15 x 4.5

V' = - 0.34 m/s

negative sign indicates direction towards left since we assumed positive direction towards right .

9.

M = mass of each railroad car = 25000 kg

V1 = velocity of single railroad car = 3.35 m/s

V2 = velocity of two coupled railroad car = 1.20 m/s

V= velocity of three coupled cars

using conservation of momentum

MV1 + (2M) V2 = (3M) V

3.35 + 2 (1.20) = 3 V

V = 1.92 m/s

b)

KE lost = (0.5) (MV21 + (2M) V22 - (3M) V2)

KE lost = (0.5) M (V21 + 2 V22 - 3 V2)

KE lost = (0.5) (25000) ((3.35)2 + 2 (1.20)2 - 3 (1.92)2) = 3.8 x 104 J

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