A 48-kg girl stands on a 8-kg wagon holding two 19.5-kg weights. She throws the weights horizontally off the back of the wagon at a speed of 9.0 m/s relative to herself.
- Assuming that the wagon was at rest initially, what is the speed of the girl after she throws both weights at the same time?
- Assuming that the wagon was at rest initially, what is the speed with which the girl will move after she throws the weights one at a time, each with a speed of 9.0 m/s relative to herself?
thankyou
Ah, I just did this about an hour ago.
I did the hard part. I assume you can do the both at the same time part.
To solve these problems, we can apply the principle of conservation of momentum. The total initial momentum of the system (girl + wagon + weights) is equal to the total final momentum of the system.
1. When the girl throws both weights at the same time:
Let's denote the velocity of the girl as Vg and the velocity of the wagon as Vw.
The total mass of the system before throwing the weights is:
M_initial = mass_girl + mass_wagon + (2 * mass_weights)
= 48 kg + 8 kg + (2 * 19.5 kg)
= 116 kg
The total initial momentum of the system is:
P_initial = M_initial * V_initial
where V_initial is the velocity of the system before throwing the weights
Since the wagon is initially at rest, the velocity of the system before throwing the weights is 0 m/s:
P_initial = 116 kg * 0 m/s
= 0 kg⋅m/s
According to the principle of conservation of momentum, the total final momentum of the system is also 0 kg⋅m/s.
After the girl throws both weights, their momentum cancels out the momentum of the girl and wagon combined. The momentum of each weight is given by:
P_weight = mass_weight * velocity_weight
The total final momentum is the sum of the momenta of the thrown-off weights:
P_final = P_weight1 + P_weight2
Since the girl throws each weight horizontally at a speed of 9.0 m/s relative to herself, the momentum of each weight is:
P_weight = mass_weight * velocity_weight
= 19.5 kg * 9.0 m/s
= 175.5 kg⋅m/s
Therefore, the total final momentum P_final is:
P_final = 175.5 kg⋅m/s + 175.5 kg⋅m/s
= 351 kg⋅m/s
Now, we apply the conservation of momentum equation:
P_initial = P_final
0 kg⋅m/s = 351 kg⋅m/s
Since the momentum is conserved, the total initial momentum and the total final momentum are equal. However, since the girl and wagon had 0 initial momentum, the total final momentum is solely determined by the thrown-off weights. Therefore, the girl's final velocity after she throws both weights at the same time is 0 m/s.
2. When the girl throws the weights one at a time:
In this case, the momentum of each weight will cancel out the momentum of the girl separately.
Similar to the previous calculation, the initial momentum of the system before throwing each weight is 0 kg⋅m/s. So, the total initial momentum is still 0 kg⋅m/s.
Since the girl throws each weight horizontally at a speed of 9.0 m/s relative to herself, the momentum of each weight is:
P_weight = 19.5 kg * 9.0 m/s
= 175.5 kg⋅m/s
When the girl throws the first weight, its momentum cancels out the initial momentum of the girl. So, the girl's final velocity after throwing the first weight is 0 m/s.
Now, for the second weight, the total final momentum will be the momentum of the second weight, which is 175.5 kg⋅m/s.
Again, we apply the conservation of momentum equation:
P_initial = P_final
0 kg⋅m/s = 175.5 kg⋅m/s
Since the only momentum change happened with the second weight, its momentum determines the final momentum of the system. Therefore, the girl's final velocity after throwing each weight one at a time, each with a speed of 9.0 m/s relative to herself, is 0 m/s.
In both cases, the girl's velocity does not change after throwing the weights.