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 ϕ south of east (as indicated in the figure). After the collision, the two-car system travels at speed vfinal at an angle θ east of north. (Figure 1)

Find the speed vfinal of the joined cars after the collision.

Express your answer in terms of v and ϕ.

Answer 1

Concepts and reason

The given problem can be solved by using the expression of law of conservation of linear momentum.

To find the final velocity, the speed of the joined of two objects after collision, use the law of conservation of linear momentum. In the given problem, the mass of the two objects are equal.

Fundamentals

The expression of law of conservation of linear momentum can be expressed as follows,

${m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}$

Here, ${m_1}$ and ${m_2}$ are the masses of the two objects, ${u_1}$ and ${u_2}$ are the initial velocities before collision, ${v_1}$ and ${v_2}$ are the velocities of the objects after collision.

Consider the diagram as follows:

Calculate initial velocity of the first car is along $+ y$ axis.

$\begin{array}{l}\\{u_{1x}} = 0\\\\{u_{1y}} = 2v\hat j\\\end{array}$

The initial velocity of the second car is $v$ . $\varphi$ is the angle made by second car with $+ x$ axis.

Divide the velocity of second car into components.

$\begin{array}{l}\\{u_{2x}} = v\cos \varphi \\\\{u_{2y}} = v\sin \varphi \left( { - \hat j} \right)\\\end{array}$

After the collision, the both the masses of the object are combined to move with mass $2m$ .

Let the final velocity of the combination be ${v_{f{\rm{inal}}}}$ .

The angle made by the final combination with $+ y$ axis is $\theta$ .

Resolve the combined velocity into two components.

$\begin{array}{c}\\{v_{{\rm{final}}x}} = {v_{final}}\sin \theta \left( {\hat i} \right)\\\\{v_{{\rm{final}}y}} = {v_{final}}\cos \theta \left( {\hat j} \right)\\\end{array}$

The equation from the law of conservation of linear net momentum along $x$ axis can be expressed as follows,

${m_1}{u_{1x}} + {m_2}{u_{2x}} = 2m{v_{{\rm{final}}x}}$

Substitute $0$ for ${u_{1x}}$ , $v\cos \varphi$ for ${u_{2x}}$ , ${v_{final}}\sin \theta \left( {\hat i} \right)$ for ${v_{{\rm{final}}x}}$ in the expression of law of conservation of linear momentum.

${m_1}\left( 0 \right) + {m_2}\left( {v\cos \varphi } \right) = 2m\left( {{v_{final}}\sin \theta \left( {\hat i} \right)} \right)$

Further solve,

$v\cos \varphi = 2m\left( {{v_{final}}\sin \theta \left( {\hat i} \right)} \right)$ …… (1)

The equation of the net momentum along $y$ axis can be expressed as follows,

${m_1}{u_{1y}} + {m_2}{u_{2y}} = 2m{v_{{\rm{finaly}}}}$

Substitute $2v$ for ${u_{1y}}$ , $v\sin \varphi \left( { - \hat j} \right)$ for ${u_{2y}}$ and ${v_{final}}\cos \theta \left( {\hat j} \right)$ for ${v_{{\rm{finaly}}}}$ in the expression of conservation of linear momentum.

${m_1}\left( {2v} \right) + {m_2}\left( {v\sin \varphi \left( { - \hat j} \right)} \right) = 2m\left( {{v_{final}}\cos \theta \left( {\hat j} \right)} \right)$

Simplify the equation,

$2v - v\sin \varphi = 2{v_{{\rm{final}}}}\cos \theta$ …… (2)

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

${\left( {v\cos \theta } \right)^2} + {\left( {2v - v\sin \varphi } \right)^2} = {\left( {2{v_{{\rm{final}}}}} \right)^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)$

From the trigonometry identity,

${\sin ^2}\theta + {\cos ^2}\theta = 1$

Further solve,

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

Ans:

The expression of final velocity, ${v_{{\rm{final}}}}$ is $\frac{v}{2}\left( {\sqrt {\left( {5 - 4\sin \varphi } \right)} } \right)$ .

Answer 2

applyina law of conservation of momentum along x and y-direction,
along y-direction,
m*(2v)-m*(vsinf)=(m+m)Vfinalcos?
v(2-sinf)=2Vfianlcos?----->A
along x-direction,
mvcosf=(m+m)Vfinalsin?
vcosf=2Vfinalsin?--->B
squaring and adding A and B
v^2(2-sinf)^2+v^2cos^2f =4Vfinal^2
Vfinal=(v/2)*v[(2-sinf)^2+cos^2f]
=(v/2)*(v(5-4sinfcosf))

Answer 3

applyina law of conservation of momentum along x and y-direction,
along y-direction,
m*(2v)-m*(vsinφ)=(m+m)Vfinalcosθ

v(2-sinφ)=2Vfianlcosθ----->A

along x-direction,

mvcosφ=(m+m)Vfinalsinθ

vcosφ=2Vfinalsinθ--->B

squaring and adding A and B

v^2(2-sinφ)^2+v^2cos^2φ =4Vfinal^2

Vfinal=(v/2)*√[(2-sinφ)^2+cos^2φ]

=(v/2)*(√(5-2sinφcosφ))

Answer 4

Giventhat the mass of the each car ism-----------------------------------------------------------------Apply conservation of momentum inhorizontal direction thenmvcosφ = 2mv_final*sinθvcosφ =2v_final*sinθθ= sin^-1 (vcosφ/2v_final)=Apply conservationof momentum inverticle direction then2mv =2mv_final*cos θ - mv sinφφ = sin^-1(2v_final*cosθ - 2v ) / v

Answer 5

Here the velocity offirst car alongx axis ls zero Velocity of first car along y axis is 2vj) Here the velocity of second car along x axis is vcos0i) Velocity of first car along y axis is vsino(-) Now x component offinal velocity Applying conservation of momentumalong xaxis we get

vcos φ_Ysax sine (1) Applying conservation of momentum along y axis we get Squaring and adding (1) and (2) fa4