a man of mass 80kg standing on a stationary frictionless trolley of mass 60kg has in his hands a parcel of mass 10 kg.He throws the parcel horizontally from the rear of the trolley at 4 m.s-1 in relation to the ground.
2.1 what is the total momentum of the system before the parcel is thrown?
2.2 what is the total momentum of the system after the parcel is thrown?
2.3 calculate the speed with which the man and the trolley move after the parcel has been thrown
2.1 0
2.2 0
2.3 The man+trolley momentum is equal and opposite to the thrown parcel's momentum.
(80+60)V + 10x4 = 0
Solve for V
A negative answer means it moves in the opposite direction from the package.
V=180
To solve this problem, we'll need to calculate the total momentum before the parcel is thrown, the total momentum after the parcel is thrown, and the speed of the man and the trolley after the parcel is thrown.
2.1 To find the total momentum of the system before the parcel is thrown, we need to calculate the momentum of the man and the trolley. The momentum of an object is given by the product of its mass and velocity.
Let's denote the mass of the man as Mm, the mass of the trolley as Mt, and the mass of the parcel as Mp.
Therefore, Mm = 80 kg (mass of the man)
Mt = 60 kg (mass of the trolley)
Mp = 10 kg (mass of the parcel)
Since the trolley is stationary before the parcel is thrown, its velocity is 0 m/s.
The man is also not moving horizontally before the parcel is thrown, so his velocity is also 0 m/s.
The total momentum of the system before the parcel is thrown is the sum of the momentum of the man and the momentum of the trolley:
Momentum of the man (Pm) = Mm × Vm
= 80 kg × 0 m/s
= 0 kg·m/s
Momentum of the trolley (Pt) = Mt × Vt
= 60 kg × 0 m/s
= 0 kg·m/s
Total momentum before the parcel is thrown = Pm + Pt
= 0 kg·m/s + 0 kg·m/s
= 0 kg·m/s
Therefore, the total momentum of the system before the parcel is thrown is 0 kg·m/s.
2.2 To find the total momentum after the parcel is thrown, we need to calculate the momentum of the man and the trolley again, but this time taking into account the velocity of the parcel.
Given that the parcel is thrown horizontally from the rear of the trolley at 4 m/s in relation to the ground, the velocity of the parcel (Vp) is 4 m/s.
The momentum of the parcel (Pp) = Mp × Vp
= 10 kg × 4 m/s
= 40 kg·m/s
Therefore, the total momentum after the parcel is thrown is the sum of the momentum of the man, the momentum of the trolley, and the momentum of the parcel:
Total momentum after the parcel is thrown = Pm + Pt + Pp
= 0 kg·m/s + 0 kg·m/s + 40 kg·m/s
= 40 kg·m/s
So, the total momentum of the system after the parcel is thrown is 40 kg·m/s.
2.3 To calculate the speed with which the man and the trolley move after the parcel has been thrown, we need to use the principle of conservation of momentum. According to this principle, the total momentum before an event is equal to the total momentum after the event.
Total momentum before the parcel is thrown = Total momentum after the parcel is thrown
0 kg·m/s = (Mm + Mt) × V
Rearranging the equation, we get:
(Mm + Mt) × V = 0 kg·m/s
Substituting the given values, we have:
(80 kg + 60 kg) × V = 0 kg·m/s
140 kg × V = 0 kg·m/s
V = 0 kg·m/s / 140 kg
V = 0 m/s
Therefore, the man and the trolley do not move horizontally after the parcel has been thrown, resulting in a speed of 0 m/s.
To answer these questions, we need to understand the concept of momentum, which is the product of an object's mass and its velocity.
1. Total momentum before the parcel is thrown:
To calculate the total momentum of the system before the parcel is thrown, we need to consider both the man and the trolley. The formula for momentum is:
Momentum = mass × velocity
The man's momentum is given by:
Momentum_man = mass_man × velocity_man
mass_man = 80 kg (given)
velocity_man = 0 m/s (since the man is standing on a stationary trolley)
Momentum_man = 80 kg × 0 m/s = 0 kg·m/s
The trolley's momentum is given by:
Momentum_trolley = mass_trolley × velocity_trolley
mass_trolley = 60 kg (given)
velocity_trolley = 0 m/s (since the trolley is stationary)
Momentum_trolley = 60 kg × 0 m/s = 0 kg·m/s
The total momentum before the parcel is thrown is the sum of the man's momentum and the trolley's momentum:
Total momentum_before = Momentum_man + Momentum_trolley
= 0 kg·m/s + 0 kg·m/s
= 0 kg·m/s
Therefore, the total momentum of the system before the parcel is thrown is 0 kg·m/s.
2. Total momentum after the parcel is thrown:
Once the man throws the parcel, it separates from the system (man and trolley) and becomes its own independent object. So, we only need to consider the momentum of the man and the trolley after the parcel is thrown.
The man's momentum after throwing the parcel is still given by:
Momentum_man = mass_man × velocity_man
mass_man = 80 kg (given)
velocity_man = 0 m/s (since the man is still standing on the trolley)
Momentum_man = 80 kg × 0 m/s = 0 kg·m/s
The trolley's momentum after the parcel is thrown is given by:
Momentum_trolley = mass_trolley × velocity_trolley
mass_trolley = 60 kg (given)
velocity_trolley = ? (to be determined)
To find the velocity of the trolley after the parcel is thrown, we can apply the principle of conservation of momentum. According to this principle, the total momentum before an event is equal to the total momentum after the event, assuming there is no external force acting.
Total momentum_before = Total momentum_after
0 kg·m/s = Momentum_man_after + Momentum_trolley_after
Since the man's momentum after throwing the parcel is 0 kg·m/s (as calculated above), we can rewrite the equation as:
0 kg·m/s = 0 kg·m/s + Momentum_trolley_after
0 kg·m/s = Momentum_trolley_after
Since the man's momentum is 0 kg·m/s, it means the total momentum after the parcel is thrown is entirely carried by the trolley. Therefore, the momentum of the trolley after the parcel is thrown is 0 kg·m/s.
2.3. Calculating the speed of the man and the trolley after the parcel is thrown:
To calculate the speed with which the man and the trolley move after the parcel has been thrown, we need to know the mass of each object. The mass of the man is 80 kg, and the mass of the trolley is 60 kg (both given).
The total momentum after the parcel is thrown is the sum of the man's momentum and the trolley's momentum:
Total momentum_after = Momentum_man_after + Momentum_trolley_after
Since the man's momentum is 0 kg·m/s and the trolley's momentum is 0 kg·m/s (as calculated above), the total momentum after the parcel is thrown is also 0 kg·m/s.
Since momentum is equal to mass times velocity, we can rewrite the equation as:
0 kg·m/s = mass_man × velocity_man_after + mass_trolley × velocity_trolley_after
0 kg·m/s = 80 kg × velocity_man_after + 60 kg × velocity_trolley_after
Since the man and the trolley are connected and share the same velocity after the parcel is thrown, we can assume their velocities are equal:
0 kg·m/s = 80 kg × velocity_after + 60 kg × velocity_after
0 kg·m/s = (80 kg + 60 kg) × velocity_after
0 kg·m/s = 140 kg × velocity_after
Therefore, velocity_after = 0 m/s.
The speed with which the man and the trolley move after the parcel has been thrown is 0 m/s. They come to rest because there are no external forces acting on them to keep them in motion, such as friction or propulsion.