Question

Two cars, both of mass m, collide and stick together. Prior to the collision, one car had been traveling north at speed 2v, while the second was traveling at speed v at an angle phi south of east (asindicated in the figure). After the collision, the two-car system travels at speed v_final at an angle theta east of north.

Part A
Find the speed v_final of the joined cars after the collision.
Express your answer in terms of v and phi.

Part B
What is the angle theta with respect to north made by the velocity vector of the two cars after the collision?
Express your answer in terms of phi. Your answer should contain an inverse trigonometric function.

Answer 1

1) m₁v₁+m₂v₂=(m₁+m₂)v conservation of momentumv= (m1v1 + m2v2)/m1+m2basically, you use conservation of momentumm1v1+m2v2=m1v +m2vsince this is an inelastic collision, the objects will havethe same velocities, so all you have to do is combine themasses

Answer 2

Concepts and reason

The concept required to solve the problem is conservation of momentum.

First, find the initial momentum of both cars in horizontal as well as vertical direction.

Then, find the final momentum of the system of cars in horizontal and vertical direction.

Finally, equate the initial and final momentum and find the angle.

Fundamentals

The expression for the momentum of an object of mass m and moving with a velocity v is as follows:

$p = mv$

Here, m is the mass and v is the velocity.

The momentum of a system is said to be conserved if there is no external force acting on the system. According to the conservation of momentum, the initial momentum ( ${p_{\rm{i}}}$ ) is equal to the final momentum ( ${p_{\rm{f}}}$ ) of the system.

${p_{\rm{i}}} = {p_{\rm{f}}}$

(a)

‎The figure 1 represents the initial and final momentums of the cars before and after the collision.

According to the conservation of momentum, the initial momentum ( ${p_{\rm{i}}}$ ) is equal to the final momentum ( ${p_{\rm{f}}}$ ) of the system.

${p_{\rm{i}}} = {p_{\rm{f}}}$

The total initial momentum in the vertical direction is as follows:

${p_{{\rm{iv}}}} = {p_{1{\rm{v}}}} + {p_{2{\rm{v}}}}$

Substitute $- mv\sin \phi$ for ${p_{{\rm{1v}}}}$ and 2mv for ${p_{2{\rm{v}}}}$ in the above expression.

${p_{{\rm{iv}}}} = - mv\sin \phi + 2mv$

The total initial momentum in the horizontal direction is as follows:

${p_{{\rm{ih}}}} = {p_{1h}}$

Substitute $mv\cos \phi$ for ${p_{1{\rm{h}}}}$ in the above expression.

${p_{{\rm{ih}}}} = mv\cos \phi$

Apply the conservation of momentum in vertical direction.

${p_{{\rm{iv}}}} = {p_{{\rm{fv}}}}$

Substitute for ${p_{{\rm{iv}}}}$ and $2m{v_{\rm{f}}}\cos \theta$ for ${p_{{\rm{fv}}}}$ in the above expression.

$\begin{array}{c}\\2mv - mv\sin \phi = 2m{v_{\rm{f}}}\cos \theta \\\\2v - v\sin \phi = 2{v_{\rm{f}}}\cos \theta \\\end{array}$

Therefore, in the vertical direction,

$2v - v\sin \phi = 2{v_{\rm{f}}}\cos \theta$ ……. (1)

Apply the conservation of momentum in horizontal direction.

${p_{{\rm{ih}}}} = {p_{{\rm{fh}}}}$

Substitute $mv\cos \phi$ for ${p_{{\rm{ih}}}}$ and $2m{v_{\rm{f}}}\sin \theta$ for ${p_{{\rm{fh}}}}$ in the above expression.

$\begin{array}{c}\\mv\cos \phi = 2m{v_{\rm{f}}}\sin \theta \\\\v\cos \phi = 2{v_{\rm{f}}}\sin \theta \\\end{array}$

Therefore, in the horizontal direction,

$v\cos \phi = 2{v_{\rm{f}}}\sin \theta$ …… (2)

Squaring and adding the equation (1) and (2).

$\begin{array}{c}\\{\left( {2{v_{\rm{f}}}} \right)^2} = 4{v^2} + {v^2}{\sin ^2}\phi - 4{v^2}{\sin ^2}\phi + {v^2}{\cos ^2}\phi \\\\ = 4{v^2} + {v^2} - 4{v^2}\sin \phi \\\\ = 5{v^2} - 4{v^2}\sin \phi \\\end{array}$

Solve for ${v_{\rm{f}}}$ .

${v_{\rm{f}}} = \frac{v}{2}\sqrt {5 - 4\sin \phi }$

According to the conservation of momentum, the initial momentum ( ${p_{\rm{i}}}$ ) is equal to the final momentum ( ${p_{\rm{f}}}$ ) of the system.

${p_{\rm{i}}} = {p_{\rm{f}}}$

The total initial momentum in the vertical direction is as follows:

${p_{{\rm{iv}}}} = {p_{1{\rm{v}}}} + {p_{2{\rm{v}}}}$

Substitute $- mv\sin \phi$ for ${p_{{\rm{1v}}}}$ and 2mv for ${p_{2{\rm{v}}}}$ in the above expression.

${p_{{\rm{iv}}}} = - mv\sin \phi + 2mv$

The total initial momentum in the horizontal direction is as follows:

${p_{{\rm{ih}}}} = {p_{1h}}$

Substitute $mv\cos \phi$ for ${p_{1{\rm{h}}}}$ in the above expression.

${p_{{\rm{ih}}}} = mv\cos \phi$

Apply the conservation of momentum in vertical direction.

${p_{{\rm{iv}}}} = {p_{{\rm{fv}}}}$

Substitute for ${p_{{\rm{iv}}}}$ and $2mv\cos \theta$ for ${p_{{\rm{fv}}}}$ in the above expression.

$\begin{array}{c}\\ - mv\sin \phi + 2mv = 2mv\cos \theta \\\\2\cos \theta = 2 - \sin \phi \\\end{array}$

Apply the conservation of momentum in horizontal direction.

${p_{{\rm{ih}}}} = {p_{{\rm{fh}}}}$

Substitute $mv\cos \phi$ for ${p_{{\rm{ih}}}}$ and $2mv\sin \theta$ for ${p_{{\rm{fh}}}}$ in the above expression.

$\begin{array}{c}\\mv\cos \phi = 2mv\sin \theta \\\\2\sin \theta = \cos \phi \\\end{array}$

Divide equation $2\sin \theta = \cos \phi$ by $2\cos \theta = 2 - \sin \phi$ .

$\begin{array}{c}\\\frac{{2\sin \theta }}{{2\cos \theta }} = \frac{{\cos \phi }}{{2 - \sin \phi }}\\\\\tan \theta = \frac{{\cos \phi }}{{2 - \sin \phi }}\\\\\theta = {\tan ^{ - 1}}\left( {\frac{{\cos \phi }}{{2 - \sin \phi }}} \right)\\\end{array}$

Ans: Part A

The final velocity of the car after collision is equal to $\frac{v}{2}\sqrt {5 - 4\sin \phi }$ .