A 60kg man standing on a stationary 40kg boat throws a 0.2kg ball with a velocity of m/s. Assuming there no friction between the man and the boat, what is the speed of the boat after man throws the ball? The velocity is 50m/s how to can find?
Let's use the conservation of momentum principle here. The total momentum before the man throws the ball is zero, as everything is stationary. When the man throws the ball, the total momentum should still be conserved and equal to the initial total momentum, i.e., zero.
Initial total momentum = Final total momentum
0 = momentum of man + momentum of boat + momentum of ball
Initially, the momentum of the man, boat, and ball is zero because they are stationary. When man throws the ball, the final momentums are:
momentum of man = (mass of man) × (velocity of man)
momentum of boat = (mass of boat) × (velocity of boat)
momentum of ball = (mass of ball) × (velocity of ball)
So our equation becomes:
0 = (60 kg) × (velocity of man) + (40 kg) × (velocity of boat) + (0.2 kg) × (50 m/s)
Now we need to find the velocity of the boat. We know that the man and the boat move with the same velocity in opposite directions because there is no external horizontal force acting on the system.
Let's represent the velocity of the man and the boat as positive and negative values (since they are moving in opposite directions). Let the velocity of the man be 'v', then the velocity of the boat will be '-v.'
0 = (60 kg) × v + (40 kg) × (-v) + (0.2 kg) × (50 m/s)
Now we can solve for 'v':
60v - 40v = 10
20v = 10
v = 10/20
v = 0.5 m/s
So, the velocity of the boat is -0.5 m/s, which means the boat moves in the opposite direction of the thrown ball with a speed of 0.5 m/s.