A 60kg man standing on a stationary 40kg boat throws a 0.2kg ball with a velocity of m/s. Assuming there is no friction between the man and the boat, what is the speed of the boat after the man throws the ball?
So the Pa=-Pb
At first the intial was zero so
0=(60+40)(v)+(0.2)(1)
100v=0.2
100v/100=0.2/100
☝️ Cancel each other
V=0.2/100
At least this is my thoughts if you any other ideas share me
zackded
The speed is 50m/s
Well, it seems like we have a ball playin' boat games here! Let's do some calculations to figure out how fast that boat will go.
Firstly, we need to calculate the initial momentum of the system. The momentum of the man is given by the equation p = m*v, where the mass of the man is 60 kg and the velocity of the ball is just a mystery we're trying to solve!
Now here comes the tricky part: when the man throws the ball, the momentum of the system remains constant. So, we can write the conservation of momentum equation as:
(m_man + m_boat)*v_initial = (m_man + m_boat)*v_final
Breaking it down, we know that the initial velocity of the ball is 0 m/s, and since the man and boat are initially stationary, we can say that v_initial = 0.
After the throwing extravaganza, the ball is no longer in the man's hands, which means the final mass of the system is just the sum of the masses of the man and the boat.
Now, let's plug some numbers in to find the speed of the boat, which will be equal to the final velocity of the system:
(60 kg + 40 kg)*0 = (60 kg + 40 kg)*v_final
0 = 100 kg * v_final
Dividing both sides by 100 kg gives us:
v_final = 0 m/s
So, after all that throwing-fun, it looks like the boat remains at a standstill. Maybe they should consider calling some dolphins for a tow next time!
To solve this problem, we can apply the principle of conservation of momentum. According to this principle, the total momentum before an event is equal to the total momentum after the event, as long as there are no external forces acting on the system.
Before the man throws the ball, the total momentum of the system (man + boat + ball) is zero because everything is stationary.
After the man throws the ball, the total momentum of the system will still be zero because there are no external forces acting on the system. However, now the ball has some momentum in one direction, so the boat should have an equal but opposite momentum to maintain the total momentum at zero.
Let's denote the velocity of the ball as v_ball and the velocity of the boat as v_boat.
The momentum of an object can be calculated using the formula momentum = mass × velocity.
The initial momentum of the system will be:
Initial momentum = (mass of the man × velocity of the man) + (mass of the boat × velocity of the boat) + (mass of the ball × velocity of the ball)
Since everything is initially stationary, the velocity of the man and the velocity of the boat are both zero. Therefore:
Initial momentum = (60 kg × 0 m/s) + (40 kg × 0 m/s) + (0.2 kg × v_ball)
Since the total momentum of the system remains zero, the final momentum after the event will also be zero. Therefore:
Final momentum = (60 kg × 0 m/s) + (40 kg × v_boat) + (0.2 kg × 0 m/s)
Now we can set the initial and final momenta equal to each other and solve for v_boat:
(60 kg × 0 m/s) + (40 kg × 0 m/s) + (0.2 kg × v_ball) = (60 kg × 0 m/s) + (40 kg × v_boat) + (0.2 kg × 0 m/s)
0 + 0 + (0.2 kg × v_ball) = 0 + (40 kg × v_boat) + 0
0.2 kg × v_ball = 40 kg × v_boat
Dividing both sides of the equation by 0.2 kg:
v_ball = 200 kg × v_boat
Now we can substitute the given values into the equation to find the speed of the boat:
6 m/s = 200 kg × v_boat
Dividing both sides of the equation by 200 kg:
v_boat = 6 m/s ÷ 200 kg
v_boat = 0.03 m/s
Therefore, the speed of the boat after the man throws the ball is 0.03 m/s.
it will be 0
Momentum before = 0, so it must be zero after:
(60+40)v + (.2)....oops, you didn't put the speed of the ball after. No solution for you.