Question A 26-kg dog is running northward at 2.7 m/’s, while a 5.3 kg cat is running eastward at 3.0 m/s.
Answer: Question A. Find the total momentum of the system B. Find the magnitude of the total momentum Explanation: Momentum can be defined as the product of mass and velocity, it is a vector quantity. Momentum = mass × velocity P = mv So, since it is a vector quantity we must take note of its direction Given that, 1. 26-kg dog is running northward at 2.7 m/’s Md = 26kg Vy = 2.7m/s •j Then, it is moving in the positive y direction Then, it is momentum is Py = Md•Vy Py = 26 × 2.7 •j Py = 70.2 kgm/s •j 2. A 5.3 kg cat is running eastward at 3.0 m/s. The cat is running towards the east, then it is moving in the positive x direction Vx = 3 •i m/s Mc = 5.3kg Px = Mc•Vx Px = 5.3 × 3 •i Px = 15.9 •i kgm/s Then, Total momentum is equal to P = Px + Py P = 15.9 •i + 70.2 •j kgm/s B. Magnitude of the total momentum Magnitude of a vector quantity x = a•i + b•j is generally given as X = √(a²+b²) Direction <X = arctan(b/a) Therefore, |P| = √(15.9²+70.2²) |P| = 71.98 kgm/s Direction <P = arctan(Py/Px) =arctan(70.2/15.9) <P = 77.24° Log in to Reply
Answer:
Question
A. Find the total momentum of the system
B. Find the magnitude of the total momentum
Explanation:
Momentum can be defined as the product of mass and velocity, it is a vector quantity.
Momentum = mass × velocity
P = mv
So, since it is a vector quantity we must take note of its direction
Given that,
1. 26-kg dog is running northward at 2.7 m/’s
Md = 26kg
Vy = 2.7m/s •j
Then, it is moving in the positive y direction
Then, it is momentum is
Py = Md•Vy
Py = 26 × 2.7 •j
Py = 70.2 kgm/s •j
2. A 5.3 kg cat is running eastward at 3.0 m/s.
The cat is running towards the east, then it is moving in the positive x direction
Vx = 3 •i m/s
Mc = 5.3kg
Px = Mc•Vx
Px = 5.3 × 3 •i
Px = 15.9 •i kgm/s
Then,
Total momentum is equal to
P = Px + Py
P = 15.9 •i + 70.2 •j kgm/s
B. Magnitude of the total momentum
Magnitude of a vector quantity
x = a•i + b•j is generally given as
X = √(a²+b²)
Direction <X = arctan(b/a)
Therefore,
|P| = √(15.9²+70.2²)
|P| = 71.98 kgm/s
Direction
<P = arctan(Py/Px) =arctan(70.2/15.9)
<P = 77.24°