A boy on a 1.9 kg skateboard initially at rest
tosses a(n) 8.0 kg jug of water in the forward
direction.
If the jug has a speed of 2.7 m/s relative to
the ground and the boy and skateboard move
in the opposite direction at 0.65 m/s, find the
boy’s mass.
Answer in units of kg
reaction=action
(1.8+M).65=8*2.7
solve for M
To find the boy's mass, we can use the principle of conservation of momentum. The total momentum before the boy tosses the jug is the same as the total momentum after the toss.
The momentum of an object is calculated by multiplying its mass by its velocity. Let's denote the boy's mass as 'm' and his velocity as 'v'.
Before the toss:
The momentum of the skateboard-boy system is (1.9 kg + m) * 0.65 m/s (opposite direction).
After the toss:
The momentum of the jug is 8.0 kg * 2.7 m/s (forward direction).
The momentum of the skateboard-boy system is (1.9 kg + m - 8.0 kg) * v (opposite direction).
Now, using the principle of conservation of momentum, we can equate the total momentum before and after the toss:
(1.9 kg + m) * 0.65 m/s = (1.9 kg + m - 8.0 kg) * v
Simplifying the equation, we get:
1.235 kg + 0.65 m = (m - 6.1 kg) * v
Next, we can use the fact that the jug has a speed of 2.7 m/s relative to the ground. This means that the velocity 'v' is equal to 2.7 m/s minus the opposite velocity of the boy-skateboard system, which is -0.65 m/s. So v = 2.7 m/s + 0.65 m/s = 3.35 m/s.
Substituting this value into the equation, we have:
1.235 kg + 0.65 m = (m - 6.1 kg) * 3.35 m/s
Now we can solve for 'm':
1.235 kg + 0.65 m = 3.35 m - 20.435 kg
Moving all the 'm' terms to one side and other terms to the other side:
0.65 m - 3.35 m = -20.435 kg - 1.235 kg
-2.7 m = -21.67 kg
Finally, dividing both sides by -2.7, we get:
m = -21.67 kg / -2.7
m ≈ 8.02 kg
Therefore, the boy's mass is approximately 8.02 kg.