Identical twins, each with mass 51.0 kg, are on ice skates and at rest on a frozen lake, which may be taken as frictionless. Twin A is carrying a backpack of mass 12.0 kg. She throws it horizontally at 3.00 m/s to Twin B. Neglecting any gravity effects, what are the subsequent speeds of Twin A and Twin B?
Twin A m/s
Twin B m/s
To solve this problem, we can apply the law of conservation of momentum. The momentum before the backpack is thrown is equal to the momentum after the backpack is thrown.
The momentum of an object is given by the product of its mass and velocity:
momentum = mass * velocity
Since the twins are at rest initially, their momentum is zero. After the backpack is thrown, the total momentum of the system should still be zero, so the momentum of Twin B must be equal in magnitude but opposite in direction to the momentum of the backpack.
We can calculate the momentum of the backpack and then use it to find the velocity of Twin B.
1. Calculate the initial momentum of the backpack:
momentum_backpack_initial = mass_backpack * velocity_backpack
= 12.0 kg * 3.00 m/s
= 36.0 kg·m/s
2. Since the total momentum of the system must be zero after the backpack is thrown, the momentum of Twin B (opposite in direction to the backpack) must also be -36.0 kg·m/s.
3. Calculate the velocity of Twin B:
momentum_twin_B = mass_twin_B * velocity_twin_B
velocity_twin_B = momentum_twin_B / mass_twin_B
= -36.0 kg·m/s / 51.0 kg
So the velocity of Twin B after receiving the backpack is approximately -0.706 m/s (rounded to three decimal places).
4. Since the total momentum of the system must be zero, the momentum of Twin A must be equal in magnitude but opposite in direction to the momentum of the backpack.
5. Calculate the velocity of Twin A:
momentum_twin_A = mass_twin_A * velocity_twin_A
velocity_twin_A = momentum_twin_A / mass_twin_A
= -36.0 kg·m/s / 51.0 kg
So the velocity of Twin A after throwing the backpack is approximately -0.706 m/s (rounded to three decimal places).
Therefore, the subsequent speeds of Twin A and Twin B are:
Twin A: -0.706 m/s
Twin B: -0.706 m/s
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the backpack is thrown is equal to the total momentum after the backpack is thrown.
Before the backpack is thrown, both twins are at rest, so their initial momenta are zero. The momentum of an object is given by the product of its mass and its velocity.
The initial momentum of Twin A is given by the product of her mass (51.0 kg) and her initial velocity (0 m/s), which is equal to 0 kg*m/s.
The initial momentum of Twin B is also zero since he's at rest.
When Twin A throws the backpack to Twin B, the momentum of the system changes. The backpack has a mass of 12.0 kg and is thrown horizontally at a velocity of 3.00 m/s.
The momentum of the backpack is given by the product of its mass (12.0 kg) and its velocity (3.00 m/s), which is equal to 36.0 kg*m/s.
According to the conservation of momentum, the initial momentum of Twin A (0 kg*m/s) plus the initial momentum of the backpack (0 kg*m/s) must be equal to the final momentum of Twin A plus the final momentum of Twin B.
So, 0 kg*m/s + 0 kg*m/s = final momentum of Twin A + final momentum of Twin B
Since the momentum is a vector quantity, the final momentum of Twin A and Twin B must have opposite directions and the same magnitude, to ensure that their vector sum is zero.
Therefore, the final momentum of Twin A is equal to -36.0 kg*m/s, and the final momentum of Twin B is equal to +36.0 kg*m/s.
The final momentum of Twin A is given by the product of her mass (51.0 kg) and her final velocity (vA), so -36.0 kg*m/s = 51.0 kg * vA.
To find the final velocity of Twin A (vA), we rearrange this equation: vA = -36.0 kg*m/s / 51.0 kg.
Calculating this, we find that the final velocity of Twin A is approximately -0.706 m/s. Note that the negative sign indicates that Twin A is moving in the opposite direction to the thrown backpack.
Since the final momentum of Twin B is equal to +36.0 kg*m/s, we can find his final velocity (vB) using the equation +36.0 kg*m/s = 51.0 kg * vB.
Rearranging this equation, we find that the final velocity of Twin B (vB) is approximately +0.706 m/s. It has the same magnitude as Twin A's final velocity, but is in the opposite direction.
So, the subsequent speeds of Twin A and Twin B are -0.706 m/s and +0.706 m/s, respectively.