A man is in a boat out away from shore. The mass of the man, the boat and everything in it is 215 kg. The boat is not moving. The man throws and anchor with a mass of 19 kg straight out from the boat with a speed of 15 m/s. What is the resulting speed of the boat?
2007-05-30 09:57:48 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Conservation of momentum:
Initial momentum = 0
= final momentum
= mass anchor * speed of anchor
+ mass of boat-anchor * speed of boat
They give you the anchor mass and velocity. Subtract that from the boat's mass to get the mass of the boat as it recoils. Solve for the speed of the boat. The way I put in the signs, it will come out negative, because it goes the other way from the way you threw the anchor.
2007-05-30 10:02:37 · answer #1 · answered by Anonymous · 0⤊ 0⤋
momentum is equal to mass * velocity. momentum also must be conserved, so in this case your total momentum must be 0 (since the boat is not moving initially). the momentum of the anchor is 19kg*15m/s = 285 = 215kg*Vm/s so V = 285/215 = 1.33 m/s
2007-05-30 17:03:55 · answer #2 · answered by j 3 · 0⤊ 0⤋
the starting momentum is zero.
We'll assume no resistance to motion.
19*15+(215-19)*vb=0
where vb is the velocity of the boat
vb=-(19*15)/(215-19)
vb is -1.45 m/s, the sign indicates the boat will move in the opposite direction to the anchor
j
2007-05-30 17:01:11 · answer #3 · answered by odu83 7 · 0⤊ 1⤋