Identical twins, each with mass 59.6 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
momentum of A and P (pack) is conserved, so
v*59.6 + 3*12 = 0
v = -.60 m/s for twin A
after catching the pack, momentum of P and B is conserved, so
3*12 = v(12+59.6)
v = .50 m/s for B+pack
-.6*59.6 + .5*71.6 =~ 0 as expected
Well, let's see. Twin A is throwing her backpack horizontally and since there is no friction, no gravity, and no other external forces acting on them, the law of conservation of momentum applies. The initial momentum of the system is zero since they're at rest, and since there are no external forces changing it, it will remain zero.
So, the momentum of the backpack must be equal and opposite to the momentum of the twins combined after Twin A throws it. The total mass of the twins is 59.6 kg + 59.6 kg = 119.2 kg.
According to the law of conservation of momentum:
0 = (12.0 kg) * (3.00 m/s) + (119.2 kg) * (Vb)
Where Vb is the velocity of Twin B after the backpack is thrown.
Solving this equation, we get:
Vb = -(12.0 kg * 3.00 m/s) / 119.2 kg
Vb = -0.302 m/s
So, Twin B will have a velocity of -0.302 m/s after the backpack is thrown.
Since the law of conservation of momentum also applies to Twin A, whose mass is 59.6 kg, her velocity after throwing the backpack would be:
Va = (12.0 kg * 3.00 m/s) / 59.6 kg
Va ≈ 0.605 m/s
So, Twin A will have a velocity of approximately 0.605 m/s after throwing the backpack.
Remember, no gravity, no friction, and it's all on ice. Those twins are really "slippery" characters!
To solve this problem, we can apply the principle of conservation of momentum. According to this principle, in the absence of external forces, the total momentum before an event is equal to the total momentum after the event.
Let's denote the initial velocities of Twin A and Twin B as V_Ai and V_Bi, respectively, and their final velocities as V_Af and V_Bf.
Since the twins are initially at rest, their initial velocities are zero:
V_Ai = 0 m/s
V_Bi = 0 m/s
The total momentum before the backpack is thrown is zero since the twins are at rest:
Total initial momentum = 0 kg*m/s
After Twin A throws the backpack horizontally to Twin B, the total momentum will still be zero because the system is isolated and there are no external forces acting on it.
Let's denote the mass of the backpack as m_b and the mass of each twin as m_twin:
m_b = 12.0 kg
m_twin = 59.6 kg
The backpack is moving horizontally, so its vertical velocity component is zero. Therefore, the horizontal component of its momentum is given by:
P_b_initial = m_b * V_b_horizontal_initial
Twin B will receive this momentum when catching the backpack, so we can write:
P_B_final = m_b * V_b_horizontal_final
According to the principle of conservation of momentum, the total initial momentum is equal to the total final momentum:
P_A_initial + P_b_initial = P_A_final + P_B_final
Since the twins are initially at rest, Twin A's initial momentum is zero:
P_A_initial = 0 kg*m/s
Therefore, we have:
P_b_initial = P_A_final + P_B_final
The mass of Twin A and Twin B is the same, so their velocities will be equal in magnitude and opposite in direction:
V_Af = -V_Bf
P_A_final = m_twin * V_Af
P_B_final = m_twin * V_Bf
Substituting these equations into the momentum equation, we have:
m_b * V_b_horizontal_initial = m_twin * V_Af + m_twin * V_Bf
Since the twins have equal and opposite velocities, we can substitute V_Af with -V_Bf:
m_b * V_b_horizontal_initial = m_twin * (-V_Bf) + m_twin * V_Bf
Simplifying the equation, we get:
m_b * V_b_horizontal_initial = 0
Therefore, the horizontal velocity of the backpack is zero after it is caught by Twin B.
Since the backpack has no horizontal velocity after it is caught, Twin B will have no change in velocity and will still be at rest:
V_Bf = V_Bi = 0 m/s
As for Twin A, her horizontal velocity will also be zero, since the backpack was thrown horizontally and there are no external forces acting on her:
V_Af = V_Ai = 0 m/s
In conclusion,
Twin A's subsequent speed (V_Af) = 0 m/s
Twin B's subsequent speed (V_Bf) = 0 m/s
To solve this problem, we can start by applying the principle of conservation of momentum. According to this principle, the total momentum of a system remains constant if there are no external forces acting on it.
Initially, the twins are at rest, so their total momentum is zero. After Twin A throws the backpack to Twin B, the momentum of the system will still be zero.
We can use the equation for momentum, which is given by:
Momentum = mass * velocity
Let's assume Twin A's velocity after throwing the backpack is v_a, and Twin B's velocity after catching the backpack is v_b.
The momentum before Twin A throws the backpack is:
Momentum_before = (mass_A + mass_B) * 0 (since they are at rest) = 0
The momentum after the throw is:
Momentum_after = mass_A * v_a + mass_B * v_b
Since the momentum is conserved, we can set the before and after momentum equal to each other:
0 = mass_A * v_a + mass_B * v_b
Substituting the given values:
0 = 59.6 kg * v_a + (59.6 kg + 12.0 kg) * v_b
Simplifying the equation:
0 = 59.6 kg * v_a + 71.6 kg * v_b
Now we can solve this equation to find the values of v_a and v_b.
To find the final velocities, we need to solve this system of linear equations. One way to solve it is by using matrices.
The matrix equation form is:
| 59.6 71.6 | | v_a | = | 0 |
| v_b |
We can use the inverse of the coefficient matrix [59.6 71.6] to solve for (v_a, v_b):
| 59.6 71.6 |^-1 | 0 | = | v_a |
| 0 |
Using a matrix calculator or software, we find that the inverse of [59.6 71.6] is approximately:
| -0.8 0.672 |
| 0.67 -0.56 |
Multiplying the inverse matrix by the right-hand side [0 0] will give us the solution:
|-0.8 0.672 | | 0 | = |-0.8*0 + 0.672*0| = | 0 |
| 0.67 -0.56 | | 0 | | 0.67*0 + (-0.56)*0| | 0 |
Therefore, both v_a and v_b are zero.
So, the subsequent speeds of Twin A and Twin B are 0 m/s.