Inelastic Collision At An Angle

[Saendy]

Two objects of mass M (= 1 kg) each travel with identical speed (|v1| = |v2| = 3 m/s) making an angle of θ relative to the x-axis. After they collide with each other, they travel as one object of mass 2M and with a velocity v3 (|v3| = 2 m/s) in the horizontal direction.

What is the value for angle θ in radians?

I tried m1 * v1 * cos(θ) + m2 * v2 * cos(θ) = mf * vf, but that was just a blind attempt to try something.

It's a practice exam question so I know the answer, but I want to know how to get there.

 

[SammyS]

Two objects of mass M (= 1 kg) each travel with identical speed (|v1| = |v2| = 3 m/s) making an angle of θ relative to the x-axis. After they collide with each other, they travel as one object of mass 2M and with a velocity v3 (|v3| = 2 m/s) in the horizontal direction.

What is the value for angle θ in radians?

I tried m1 * v1 * cos(θ) + m2 * v2 * cos(θ) = mf * vf, but that was just a blind attempt to try something.

It's a practice exam question so I know the answer, but I want to know how to get there.

Hello Saendy. Welcome to PF !

Use conservation of momentum.

What's the momentum of the system (total momentum) before the collision?

What's the momentum of the system after the collision?

 

[Saendy]

Hi Sammy, thanks for replying!

The total momentum before should be:

Pi = (m1 * v1) + (m2 * v2) = (3*1) + (3*1) = 6

The total momentum after should be:

Pf = (2m * vf) = (2*2) = 4

 

[SammyS]

Hi Sammy, thanks for replying!

The total momentum before should be:

Pi = (m1 * v1) + (m2 * v2) = (3*1) + (3*1) = 6

Don't forget v₁ andv₂ are vectors. Momentum is too.

The total momentum after should be:

Pf = (2m * vf) = (2*2) = 4

 

[Nickie]

The value for angle θ in radians can be found by using the conservation of momentum and conservation of kinetic energy equations. First, we can set up the equations for each of these principles:

Conservation of momentum: m1*v1*cos(θ) + m2*v2*cos(θ) = mf*v3*cos(0)

Conservation of kinetic energy: (1/2)*m1*v1^2 + (1/2)*m2*v2^2 = (1/2)*mf*v3^2

Since we know the masses and velocities of the objects before and after the collision, we can plug in these values and solve for θ. We also know that the objects are traveling at the same speed, so we can set v1 = v2 = v and simplify the equations:

m1*v*cos(θ) + m2*v*cos(θ) = mf*v3

(1/2)*m1*v^2 + (1/2)*m2*v^2 = (1/2)*mf*v3^2

Now we can substitute in the values of m1, m2, mf, v, and v3:

(1/2) * (1 kg) * (3 m/s) * cos(θ) + (1/2) * (1 kg) * (3 m/s) * cos(θ) = (2 kg) * (2 m/s) * cos(0)

(1/2) * (1 kg) * (3 m/s)^2 + (1/2) * (1 kg) * (3 m/s)^2 = (1/2) * (2 kg) * (2 m/s)^2

Simplifying these equations, we get:

(3/2) * cos(θ) = 4

(9/2) * cos(θ) = 8

Dividing both equations by 2, we get:

(3/4) * cos(θ) = 2

(9/4) * cos(θ) = 4

Solving for cos(θ), we get:

cos(θ) = 8/3

cos(θ) = 4/9

Using a calculator, we can find the inverse cosine of these values to get the angle θ:

θ = 69.44 degrees or 1.21 radians

Therefore, the value for