Question

Two gliders collide on an air track moving in from opposite directions, and they bounce back. The masses of the gliders are 0.20 kg and 0.30 kg. The magnitudes of their initial velocities are 0.50 m/s and 0.40 m/s, respectively. After the collision, the magnitudes of their velocities are 0.40 m/s and 0.20 m/s, respectively. Assume the first glider is initially heading in the positive direction.The total momentum of the gliders before collision is closest to:

Answer 1

Answer:

so initial momentum is 0.22kgm/s

Explanation:

m1=0.20kg

m2=0.30kg

initial velocity of m1=u1=0.50m/s

initial velocity of m2=u2=0.40m/s

total momentum of the system before collision

Pi=m1u1+m2u2

Pi=0.20kg×0.50m/s+0.30kg×0.40m/s

Pi=0.1kgm/s+0.12kgm/s

Pi=0.22kgm/s

Answer 2

The total momentum of the gliders before the collision, with masses 0.20 kg and 0.30 kg, respectively, and initial velocities of 0.50 m/s and 0.40 m/s, respectively is -0.02 kg*m/s.  

Before the collision, the first glider is heading in the positive direction (to the right) and the second glider is moving in the negative direction (to the left).  

The total momentum before the collision is given by:

$p_{i} = m_{1}v_{1} + m_{2}(v_{2})$

Where:

m₁: is the mass of the first glider = 0.20 kg

m₂: is the mass of the second glider = 0.30 kg

v₁: is the velocity of the first glider = 0.50 m/s

v₂: is the velocity of the second glider = -0.40 m/s (minus sign because the glider is in the negative direction)

Then, the total momentum before the collision is:  

$p_{i} = (0.20*0.50 - 0.30*0.40) kg*m/s = -0.02 kg*m/s$

Therefore, the total momentum of the gliders before the collision is -0.02 kg*m/s.

Find more here:

• https://brainly.com/question/6391301?referrer=searchResults

• https://brainly.com/question/24592032?referrer=searchResults

I hope it helps you!