Confirming Velocity Direction After Inelastic Collision

[tobywashere]

Homework Statement

I can solve this problem, but I keep getting a different answer from the textbook. I just want to confirm whether I'm wrong or the textbook is wrong.

A truck of mass 2.3 x 10⁴ kg traveling at 15 m/s [51° S of W] collides with a second truck of mass 1.2 x 10⁴ kg. The collision is completely inelastic. The trucks have a common velocity of 11 m/s [35° S of W] after the collision.

Determine the initial velocity of the less massive truck.

Homework Equations

v`(m1 + m2) = v1m1 + v2m2
Cosine law
Sine law

The Attempt at a Solution

P2² = 385000² +345000² -2(385000)(345000)cos16°
P2 = 109045
V = P2 / m2
V = 109045 / 1.2 x 10⁴
V = 9.1 m/s
This part of my answer is the same as the textbook's answer. It's the direction of the velocity that differs.

(sin16°)/109045= sinθ/385000
θ = 77°
Since the momentum of the lighter second truck is in the north west quadrant:
180° - (77° + 51°) = 53°
Therefore, the velocity of the second truck is 9.1 m/s [53° N of W]
The textbook says that the velocity is [26° N of W]
Can anyone check if this is right?

 

[tobywashere]

bummmmmppppp

 

[nvn]

tobywashere: Your answer is incorrect. The textbook is correct. Keep trying.

 

[cupid.callin]

Write the momenta of the bodies before and after collision along X and Y axis
take the unknown velocity as: x(i) +y(j)

Use conservation of momentum separately along X and Y axis!

 

[nvn]

tobywashere: Hint: Your approach in post 1 seemed fairly good, except I currently think it failed because, it just so happened, you ran into the ambiguous case[/color] of the sine law. Try again.

 

[SammyS]

Homework Statement

I can solve this problem, but I keep getting a different answer from the textbook. I just want to confirm whether I'm wrong or the textbook is wrong.

A truck of mass 2.3 x 10⁴ kg traveling at 15 m/s [51° S of W] collides with a second truck of mass 1.2 x 10⁴ kg. The collision is completely inelastic. The trucks have a common velocity of 11 m/s [35° S of W] after the collision.

Determine the initial velocity of the less massive truck.

Homework Equations

v`(m1 + m2) = v1m1 + v2m2
Cosine law
Sine law

The Attempt at a Solution

P2² = 385000² +345000² -2(385000)(345000)cos16°
P2 = 109045
V = P2 / m2
V = 109045 / 1.2 x 10⁴
V = 9.1 m/s
This part of my answer is the same as the textbook's answer. It's the direction of the velocity that differs.

(sin16°)/109045= sinθ/385000
θ = 77°
Since the momentum of the lighter second truck is in the north west quadrant:
180° - (77° + 51°) = 53°
Therefore, the velocity of the second truck is 9.1 m/s [53° N of W]
The textbook says that the velocity is [26° N of W]
Can anyone check if this is right?

The problem is that sin(103°) is the same as sin(77°).

You used the Law of Cosines to find the third side (p₂) of the momentum triangle. There's no problem with that.

Then you used the Law Of Sines to find the angle opposite the longest side of the triangle. That can give two possible results, depending upon whether the triangle has an obtuse angle or not.

If you had used the Law of Sines to find the other unknown angle, the angle opposite p₁ (345000 kg·m2/s²) that angle cannot be obtuse, since p₁ is not the longest side.

Find angle ɸ, which is opposite side p₁ using Law of Sines:

(sin16°)/109045 = (sin ɸ)/345000

Then find θ by subtracting ɸ and 16° from 180°.