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
Aww... what happened?
I will post later and share what happened.... it was a disturbing experience, though....
Frame what happened as a lesson. Think out what happened, and what it means. Then, see it as an opportunity to learn, and be prepared in the future. When you are older, you can teach your kids based on your experiences, and they win from it. Perhaps even your grandchildren will learn valuable lessons from it.
I am sorry it went badly for you, but do your best to make something good from the remains.
Thanks, Bob... I will seriously think about that.... I'll post a more detailed message tomorrow letting everyone know what actually happened
To find the subsequent speeds of Twin A and Twin B, we can use the law of conservation of linear momentum. According to this principle, the total momentum before the backpack is thrown should be equal to the total momentum after the backpack is thrown.
The momentum (p) of an object is defined as the product of its mass (m) and velocity (v).
Before the backpack is thrown:
Momentum of Twin A (pA) = mass of Twin A (mA) * velocity of Twin A (vA)
Momentum of Twin B (pB) = mass of Twin B (mB) * velocity of Twin B (vB)
Since both twins are at rest initially, their velocities are 0. Therefore, the total momentum before the backpack is thrown is 0.
After the backpack is thrown:
Momentum of Twin A (pA') = (mass of Twin A without backpack) * velocity of Twin A (vA')
Momentum of Twin B (pB') = (mass of Twin B + mass of backpack) * velocity of Twin B (vB')
According to the law of conservation of linear momentum:
pA + pB = pA' + pB'
Since the total momentum before the backpack is thrown is 0, we have:
0 + 0 = pA' + pB'
Substituting the momentum expressions, we get:
0 = (mA * vA) + (mB * vB')
Now we can solve for vA' and vB':
vA' = - (mB * vB') / mA (1)
vB' = (mA * vA) / (mB + mass of backpack) (2)
Substituting the given values:
mA = 51.0 kg (mass of Twin A)
mB = 51.0 kg (mass of Twin B)
mass of backpack = 12.0 kg (mass of the backpack)
vA = 0 m/s (initial velocity of Twin A)
vB = 0 m/s (initial velocity of Twin B)
Using equation (2):
vB' = (51.0 kg * 0 m/s) / (51.0 kg + 12.0 kg)
vB' = 0 m/s
Using equation (1):
vA' = - (51.0 kg * 0 m/s) / 51.0 kg
vA' = - 0 m/s
vA' = 0 m/s
Therefore, the subsequent speeds of Twin A and Twin B are:
Twin A: 0 m/s
Twin B: 0 m/s