Identical twins, each with mass 45.5 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 2.80 m/s to Twin B. Neglecting any gravity effects, what are the subsequent speeds of Twin A and Twin B?Twin A m/sTwin B m/s
Question
Identical twins, each with mass 45.5 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 2.80 m/s to Twin B. Neglecting any gravity effects, what are the subsequent speeds of Twin A and Twin B?Twin A m/sTwin B m/s
Solution
This problem can be solved using the principle of conservation of momentum. The total momentum before the backpack is thrown is zero (since both twins are at rest), and the total momentum after the backpack is thrown must also be zero (since there are no external forces).
Let's denote the final speed of Twin A as Va and the final speed of Twin B as Vb.
Before the backpack is thrown, the total momentum (P_initial) is 0. This is because both twins and the backpack are at rest.
After the backpack is thrown, the total momentum (P_final) is the sum of the momenta of Twin A, Twin B, and the backpack. This can be expressed as:
P_final = (mass of Twin A * Va) + (mass of Twin B * Vb) + (mass of backpack * speed of backpack)
Since the total momentum must be conserved, we can set P_initial equal to P_final and solve for Va and Vb:
0 = (45.5 kg * Va) + (45.5 kg * Vb) + (12.0 kg * 2.80 m/s)
We know that Twin A throws the backpack to Twin B, so she will move in the opposite direction of the backpack. Therefore, Va will be negative and Vb will be positive.
We can solve the above equation for Va:
Va = - (12.0 kg * 2.80 m/s) / 45.5 kg = -0.738 m/s
And for Vb:
Vb = - (45.5 kg * Va) / 45.5 kg = 0.738 m/s
So, Twin A will move at 0.738 m/s in the opposite direction of the backpack, and Twin B will move at 0.738 m/s in the same direction as the backpack.
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