Identical twins, each with mass 56.2 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 3.50 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
Try using the law of conservation of linear momentum, twice. First after the backpack is thrown, and again after it is caught. Show your work if you need further help.
0.25
To find the subsequent speeds of Twin A and Twin B after Twin A throws the backpack, we can apply the principle of conservation of momentum.
According to the principle of conservation of momentum, the total momentum before the interaction is equal to the total momentum after the interaction, assuming no external forces act on the system.
The total momentum before the interaction is the sum of the momenta of Twin A and the backpack. Since they are initially at rest, their momenta are both zero.
The total momentum after the interaction is the sum of the momenta of Twin A and Twin B. Since the backpack is thrown from Twin A to Twin B horizontally, there is no vertical momentum involved. The horizontal momentum of the backpack before and after the throw is zero.
Therefore, we can write the conservation of momentum equation as:
0 + 0 = (ma * va) + (mb * vb)
Where:
- ma is the mass of Twin A (56.2 kg),
- va is the initial velocity of Twin A (0 m/s),
- mb is the mass of Twin B (56.2 kg),
- vb is the final velocity of Twin B (which we need to find).
Since the horizontal momentum of the backpack is zero before and after the throw, we can neglect it in the equation.
Now, we can solve the equation for vb:
0 = (56.2 kg * 0 m/s) + (56.2 kg * vb)
0 = 0 + (56.2 kg * vb)
0 = 56.2 kg * vb
Since the mass of Twin B is the same as Twin A, the coefficient in front of vb cancels out. Therefore, vb must be zero for the equation to hold true.
This means that after Twin A throws the backpack to Twin B horizontally, Twin B remains at rest on the ice. Therefore, the subsequent speed of Twin A is 0 m/s, and the subsequent speed of Twin B is 0 m/s.