Homework due tomorrow!!!

[Jameson]

A 2082 kg Oldsmobile traveling east on Saginaw Street at 14.3 m/s is unable to stop on the ice covered intersection for a red light at Abbott Road. The car collides with a 4070 kg truck hauling animal feed north on Abbott at 10.8 m/s. The two vehicles remain locked together after the impact. Calculate the velocity of the wreckage immediately after the impact. Give the speed for your first answer and the compass heading for your second answer. (remember, the CAPA abbreviation for degrees is deg)

I know I've already posted and I'm sorry, but I need someone to help me not so vaguely. I know about the conservation of momentum! Something just isn't adding up.

(2082*14.3)^2 + (4070*10.8)^ = 8.63

angle = 55.9

HELP ME

[daster]

Apply the principle of conservation of momentum twice, once on the x-resolute and once on the y-resolute.

m_1v_1\cos\theta+m_2v_2\cos\dot{\theta}=(m_1+m_2)v \cos\ddot{\theta}

m_1v_1\sin\theta+m_2v_2\sin\dot{\theta}=(m_1+m_2)v \sin\ddot{\theta}

Where \theta=45 and \dot{\theta}=0.

[learningphysics]

Apply the principle of conservation of momentum twice, once on the x-resolute and once on the y-resolute.

m_1v_1\cos\theta+m_2v_2\cos\dot{\theta}=(m_1+m_2)v \cos\ddot{\theta}

m_1v_1\sin\theta+m_2v_2\sin\dot{\theta}=(m_1+m_2)v \sin\ddot{\theta}

Where \theta=45 and \dot{\theta}=0.

Shouldn't the angles be 0 and 90 before collision?

[daster]

Yes, sorry. 0 and 90, not 45. :smile: