Question

Two cars, both of mass m, collide and stick together. Prior to the collision, one car had been traveling north at a speed 2v, while the second was traveling in a southeastern direction at an angle ϕ with respect to the east-west direction and at a speed v. After the collision, the two-car system travels in a northeastern direction at an angle θ with respect to the north-south direction and at a speed vfinal.
Find v_final, the speed of the joined cars after the collision.

Answer 1

Answer:$u=\frac{v}{2}\sqrt{5-4sin\phi }$

Explanation:

Given

Both cars mass is m

and solving problem in Vertical and horizontal direction

considering + y and +x to be positive and u be the final velocity of system

Conserving Momentum in Vertical direction

$m(2v)+m(-vsin\phi )=2m(ucos\theta )$

$2ucos\theta =v(2-sin\theta )$------1

Conserving momentum in x direction

$mvcos\phi =2musin\theta$-----2

squaring and adding 1 &2

$(2u)^2=(2v-vsin\phi )^2+(vcos\phi )^2$

$4u^2=4v^2+v^2-4v^2sin\phi$

$4u^2=5v^2-4v^2sin\phi$

$u=\frac{v}{2}\sqrt{5-4sin\phi }$