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° 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°