A 48-kg girl stands on a 8-kg wagon holding two 19.5-kg weights. She throws the weights horizontally off the back of the wagon at a speed of 9.0 m/s relative to herself. Assuming that the wagon was at rest initially, what is the speed with which the girl will move after she throws the weights one at a time, each with a speed of 9.0 m/s relative to herself?
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the girl throws the weights should be equal to the total momentum after she throws them.
1. Calculate the initial momentum:
- The girl's initial momentum is given by her mass multiplied by her velocity, which is 48 kg × 0 m/s (since the wagon is at rest).
- The weights are initially at rest on the wagon, so their initial momentum is 0.
2. Calculate the momentum after each weight is thrown:
- When the girl throws the first weight, she loses 19.5 kg × 9.0 m/s of momentum in the backward direction.
- The momentum of the first weight is given by its mass (19.5 kg) multiplied by its velocity (9.0 m/s) in the backward direction.
- After the first weight is thrown, the girl's momentum is negative since she loses momentum in the backward direction.
- When the girl throws the second weight, she loses another 19.5 kg × 9.0 m/s of momentum in the backward direction.
- The momentum of the second weight is given by its mass (19.5 kg) multiplied by its velocity (9.0 m/s) in the backward direction.
- After throwing the second weight, the girl's momentum is still negative since she loses more momentum in the backward direction.
3. Calculate the final momentum:
- The total momentum after throwing the weights is the sum of the girl's momentum and the momentum of the thrown weights.
4. Calculate the final velocity of the girl:
- Since momentum is defined as mass multiplied by velocity, we can rearrange the equation to solve for velocity.
- Divide the final momentum (calculated in step 3) by the total mass of the system (girl's mass and weights' masses combined) to find the final velocity.
Let's calculate the values step-by-step:
1. Initial momentum:
- Girl's initial momentum = 48 kg × 0 m/s = 0 kg·m/s
- Weights' initial momentum = 0 kg × 0 m/s = 0 kg·m/s
2. Momentum after each weight is thrown:
- Momentum after throwing the first weight = -19.5 kg × 9.0 m/s = -175.5 kg·m/s
- Momentum after throwing the second weight = -2 × 19.5 kg × 9.0 m/s = -351.0 kg·m/s
3. Total momentum after throwing the weights:
- Total momentum = Girl's momentum + Weights' momentum
= (-175.5 kg·m/s) + (-351.0 kg·m/s)
= -526.5 kg·m/s
4. Final velocity of the girl:
- Total mass of the system = Girl's mass + Weights' masses
= 48 kg + (2 × 19.5 kg)
= 87 kg
- Final velocity = Total momentum / Total mass
= -526.5 kg·m/s / 87 kg
= -6.05 m/s
Therefore, the speed with which the girl will move after throwing the weights is approximately 6.05 m/s in the backward direction.
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum of a system remains constant if no external forces are acting on it.
Initially, the girl and the wagon are at rest, so their total momentum is zero. After the girl throws each weight horizontally off the back of the wagon, the total momentum of the system remains zero because there are no external forces acting on it.
Let's denote the velocity of the girl after throwing the weights as v. To find the value of v, we can use the law of conservation of momentum.
The initial momentum of the system is given by:
Initial momentum = momentum of the girl + momentum of the wagon
Since they are at rest, the initial momentum is zero:
Initial momentum = 0
The final momentum of the system is given by:
Final momentum = momentum of the girl + momentum of the wagon + momentum of the first weight + momentum of the second weight
Since the weights are thrown in opposite directions, their momenta are opposite in direction but equal in magnitude:
Momentum of the first weight = -mass of the first weight * velocity of the first weight
Momentum of the second weight = -mass of the second weight * velocity of the second weight
Using the principle of conservation of momentum, we can set up the equation:
Initial momentum = Final momentum
0 = mass of girl * velocity of girl + mass of wagon * velocity of wagon - (mass of first weight * velocity of first weight + mass of second weight * velocity of second weight)
Substituting the given values into the equation:
0 = 48 kg * v + 8 kg * 0 - (19.5 kg * 9.0 m/s + 19.5 kg * 9.0 m/s)
Simplifying the equation:
0 = 48 kg * v - (19.5 kg * 9.0 m/s + 19.5 kg * 9.0 m/s)
0 = 48 kg * v - (2 * 19.5 kg * 9.0 m/s)
0 = 48 kg * v - 351 kg*m/s
Rearranging the equation to solve for v:
48 kg * v = 351 kg*m/s
v = 351 kg*m/s / 48 kg
v ≈ 7.31 m/s
Therefore, the speed with which the girl will move after she throws the weights one at a time is approximately 7.31 m/s.
Weight number 1:
48 + 8 + 19.5 = 75.5 kg on the wagon
19.5 kg back at 9 relative to ground and girl
so
75.5 u = 19.5 * 9
u = 2.325 forward
now
weight number 2
before throw:
momentum = 75.5 * 2.325 = 175.5
relative to ground
after throw
missile momentum back relative to ground
= 19.5 (9 -2.325 ) = 130 kg m/s back
cart and girl mass = 48+8 = 56 kg
so
56 v - 130 = momentum after
so
56 v - 130 = 175.5
56 v = 305.5
v = 5.46 meters/second