Question

A blue car with mass mc = 427 kg is moving east with a speed of vc = 20 m/s and collides with a purple truck with mass mt = 1282 kg that is moving south with a speed of vt = 13 m/s . The two collide and lock together after the collision. Let us define to the east is the +x direction and to the north is the +y direction. What is the magnitude of the initial momentum of the car?

Answer 1

Answer:8540 kg-g/s

Explanation:

Given

mass of blue car $m_c=427 kg$

velocity of blue car $v_c=20 m/s$

mass of the truck $m_t=1282 kg$

speed of truck $v_t=13 m/s$

After collision they stick and lock together

Let v be the velocity of combined system at angle \theta from vertical

Conserving momentum in east direction

$m_c\times v_c=(m_c+m_t)v\cos \theta$

$427\times 20 =1709\times v\cos \theta$------1

Conserving Momentum in Y direction

$m_t\times v_t=(m_c+m_t)v\sin \theta$

$1282\times 13 =1709\times v\sin \theta$-------2

squaring and then adding 1 & 2 we get

$(8540)^2+(16666)^2=(1709)^2\cdot v^2$

v=10.95 m/s

initial momentum of car$=427\times 20=8540 kg-m/s$