Answer:
The magnitude of the total momentum is 21.2 kg m/s and its direction is 39.5° from the x-axis.
Explanation:
Hi there!
The total momentum is calculated as the sum of the momenta of the pieces.
The momentum of each piece is calculated as follows:
p = m · v
Where:
p = momentum.
m = mass.
v = velocity.
The momentum is a vector. The 200 g-piece flies along the x-axis then, its momentum will be:
p = (m · v, 0)
p = (0.200 kg · 82.0 m/s, 0)
p = (16.4 kg m/s, 0)
The 300 g-piece flies along the y-axis. Its momentum vector will be:
p =(0, m · v)
p = (0, 0.300 kg · 45.0 m/s)
p = (0, 13.5 kg m/s)
The total momentum is the sum of each momentum:
Total momentum = (16.4 kg m/s, 0) + (0, 13.5 kg m/s)
Total momentum = (16.4 kg m/s + 0, 0 + 13.5 kg m/s)
Total momentum = (16.4 kg m/s, 13.5 kg m/s)
The magnitude of the total momentum is calculated as follows:
[tex]|p| = \sqrt{(16.4 kgm/s)^2+(13.5 kg m/s)^2}= 21.2 kg m/s[/tex]
The direction of the momentum vector is calculated using trigonometry:
cos θ = px/p
Where px is the horizontal component of the total momentum and p is the magnitude of the total momentum.
cos θ = 16.4 kg m/s / 21.2 kg m/s
θ = 39.3 (39.5° if we do not round the magnitude of the total momentum)
Then, the magnitude of the total momentum is 21.2 kg m/s and its direction is 39.5° from the x-axis.
