Question

A 1000-kg car traveling north at 15 m/s collides with a2000-kg truck traveling east at 10 m/s. The occupants, wearing seat belts, are uninjured, butthe two vehicles move away from the impact point as one. The insurance adjuster asks youto find the velocity of the wreckage just after impact. What is your answer?

Answer 1

8.33 m/s, 36.87° North of East

Step-by-step explanation:

`m_n` = Mass of car = 1000 kg

`v_n` = Velocity of car = 15 m/s

`m_e` = Mass of truck = 2000 kg

`v_e` = Velocity of truck = 10 m/s

M = Combined mass = 1000+2000 = 3000 kg

Momentum

`p_n=m_nv_n\\\Rightarrow p_n=1000* 15\\\Rightarrow p_n=15000\ kgm/s`

Momentum of car traveling East is 15000 kgm/s

`p_e=m_ev_e\\\Rightarrow p_n=2000* 10\\\Rightarrow p_n=20000\ kgm/s`

Momentum of truck traveling North is 20000 kgm/s

Angle

`\theta=tan^(-1)(p_n)/(p_e)\\\Rightarrow \theta=tan^(-1)(15000)/(20000)\\\Rightarrow \theta=36.87^(\circ)`

As the two vehicles are vectors, the resultant velocity is

`(Mv)^2=p_n^2+p_e^2\\\Rightarrow v=\sqrt{(p_n^2+p_e^2)/(M^2)}\\\Rightarrow v=\sqrt{(15000^2+20000^2)/(3000^2)}\\\Rightarrow v=8.33\ m/s`

Velocity of the two vehicles when they are locked together is 8.33 m/s and direction is 36.87° North of East