Part 1
Calculate the magnitude of the linear momentum for each of the following cases:
A proton with mass 1.67×10
−27
kg moving
with a speed of 5.24 × 10
6 m/s.
Answer in units of kg · m/s
Part 2
A 15.1 g bullet moving with a speed of
327 m/s.
Answer in units of kg · m/s
Part 3
A 74.9 kg sprinter running with a speed of
13.8 m/s .
Answer in units of kg · m/s
Part 4
Earth (mass = 5.98 × 10
24
kg) moving with
an orbital speed equal to 28803.3 m/s .
Answer in units of kg · m/s
Part 1:
To calculate the magnitude of linear momentum, we use the formula:
linear momentum = mass × velocity
Given:
mass of the proton = 1.67 × 10^(-27) kg
velocity of the proton = 5.24 × 10^6 m/s
Substituting these values into the formula, we get:
linear momentum = (1.67 × 10^(-27) kg) × (5.24 × 10^6 m/s)
linear momentum = 8.75 × 10^(-21) kg · m/s
Therefore, the magnitude of the linear momentum for the proton is 8.75 × 10^(-21) kg · m/s.
Part 2:
Given:
mass of the bullet = 15.1 g = 0.0151 kg (converted to kg)
velocity of the bullet = 327 m/s
Using the formula for linear momentum:
linear momentum = mass × velocity
Substituting the values:
linear momentum = (0.0151 kg) × (327 m/s)
linear momentum = 4.9437 kg · m/s
Therefore, the magnitude of the linear momentum for the bullet is 4.9437 kg · m/s.
Part 3:
Given:
mass of the sprinter = 74.9 kg
velocity of the sprinter = 13.8 m/s
Using the formula for linear momentum:
linear momentum = mass × velocity
Substituting the values:
linear momentum = (74.9 kg) × (13.8 m/s)
linear momentum = 1033.62 kg · m/s
Therefore, the magnitude of the linear momentum for the sprinter is 1033.62 kg · m/s.
Part 4:
Given:
mass of Earth = 5.98 × 10^24 kg
orbital speed of Earth = 28803.3 m/s
Using the formula for linear momentum:
linear momentum = mass × velocity
Substituting the values:
linear momentum = (5.98 × 10^24 kg) × (28803.3 m/s)
linear momentum = 1.722 × 10^29 kg · m/s
Therefore, the magnitude of the linear momentum for Earth is 1.722 × 10^29 kg · m/s.
Part 1:
To calculate the magnitude of linear momentum, we use the formula:
linear momentum = mass × velocity
Given:
mass of proton = 1.67×10^-27 kg
velocity = 5.24 × 10^6 m/s
linear momentum = (1.67×10^-27 kg) × (5.24 × 10^6 m/s)
linear momentum = 8.75 × 10^-21 kg · m/s
Therefore, the magnitude of the linear momentum for a proton is 8.75 × 10^-21 kg · m/s.
Part 2:
Given:
mass of bullet = 15.1 g = 0.0151 kg
velocity = 327 m/s
linear momentum = (0.0151 kg) × (327 m/s)
linear momentum = 4.9407 kg · m/s
Therefore, the magnitude of the linear momentum for the bullet is 4.9407 kg · m/s.
Part 3:
Given:
mass of sprinter = 74.9 kg
velocity = 13.8 m/s
linear momentum = (74.9 kg) × (13.8 m/s)
linear momentum = 1032.12 kg · m/s
Therefore, the magnitude of the linear momentum for the sprinter is 1032.12 kg · m/s.
Part 4:
Given:
mass of Earth = 5.98 × 10^24 kg
velocity = 28803.3 m/s
linear momentum = (5.98 × 10^24 kg) × (28803.3 m/s)
linear momentum = 1.7223 × 10^29 kg · m/s
Therefore, the magnitude of the linear momentum for Earth is 1.7223 × 10^29 kg · m/s.