How Is Momentum Conserved in a Two-Body Collision?

[kmikias]

Hi.I have some confusion on this question ...here is the question

1.A(n) 2.6 kg object moving with a speed of
7.9 m/s collides with a(n) 3 kg object mov-
ing with a velocity of 7.4 m/s in a direction
56.1768◦ from the initial direction of motion
of the 2.6 kg object.What is the speed of the two objects after
the collision if they remain stuck together?
Answer in units of m/s

here is what i did but my answer out to be wrong .
Mass * velocity + Mass * velocity =( Mass +Mass)velocity

2.6 * 7.9 + 3*7.4 *cos56.1768 = (2.6 +3)v
20.54 +12.357 = 5.6v
so velocity = 5.8745 m/s

thank you.

 

[n00bhaus3r]

You're only considering the component of the velocity in the direction of the first object. That's why you did cos(56.1768). Don't forget to factor in the other component, namely, sin(56.1768). You should get the right answer if you do.

 

[thomate1]

Hi there,

I understand your confusion with this question. The key concept to understand here is the conservation of momentum. This means that the total momentum of the system (both objects) before the collision is equal to the total momentum after the collision. In other words, momentum is conserved.

In order to solve this problem, we can use the formula for momentum: p = mv, where p is momentum, m is mass, and v is velocity.

Before the collision, the momentum of the 2.6 kg object is:
p1 = (2.6 kg)*(7.9 m/s) = 20.54 kg*m/s

The momentum of the 3 kg object is:
p2 = (3 kg)*(7.4 m/s*cos56.1768) = 12.357 kg*m/s

The total momentum before the collision is:
p = p1 + p2 = 20.54 + 12.357 = 32.897 kg*m/s

After the collision, the two objects stick together and move with the same velocity, so we can use the same formula to find the final velocity:
p = (5.6 kg)*v

Solving for v, we get:
v = p/(5.6 kg) = 32.897 kg*m/s / (5.6 kg) = 5.8745 m/s

So the final velocity of the two objects after the collision is 5.8745 m/s. I hope this helps clarify any confusion you had with this problem. Remember to always consider the conservation of momentum when dealing with collisions. Keep up the good work!