CameraIcon

SearchIcon

MyQuestionIcon

205

You visited us 205 times! Enjoying our articles? Unlock Full Access!

Definition of a Projectile

A mass m 1 ...

Question

A mass $m_{1}$ with initial speed $v_{0}$ in the positive $x$ -direction collides with a mass $m_{2} = 2 m_{1}$ which is initially at rest at the origin, as shown in figure. After the collision $m_{1}$ moves off with speed $v_{1} = v_{0} / 2$ in the negative $y$ - direction, and $m_{2}$ moves off with speed $v_{2}$ at angle $θ$ . If the energy change due the collision is $\frac{m {v_{0}}^{2}}{10 + x}$ . Find $x$ .

Open in App

Solution

By the law of conservation of linear momentum,
along x-axis ,
total momentum before collision=total momentum after collision
$m_{1} v_{0} + 0 = m_{1} \times 0 + m_{2} v_{2} cos θ$ ,
or $m_{1} v_{0} = 2 m_{1} v_{2} cos θ$ , (given $2 m_{2} = 2 m_{1}$ ) ,
or $v_{0} = 2 v_{2} cos θ$ .......eq1 ,
along y-axis ,
total momentum before collision=total momentum after collision
$0 = - m_{1} v_{1} + m_{2} v_{2} sin θ$ ,
or $0 = - m_{1} v_{0} / 2 + 2 m_{1} v_{2} sin θ$ , (given $v_{1} = v_{0} / 2$ ) ,
or $v_{0} / 4 = v_{2} sin θ$ .......eq2,
by squarring and adding eq1 and 2 , we get
$v_{2} = \frac{\sqrt{5}}{4} v_{0}$ ,
Now , change in kinetic energy will be ,
$Δ k =$ kinetic energy after collision-kinetic energy before collision ,
or $Δ k = 1 / 2 m_{1} v_{1}^{2} + 1 / 2 m_{2} v_{2}^{2} - 1 / 2 m_{1} v_{0}^{2} + 0$ ,
or $Δ k = 1 / 2 m_{1} (v_{0}^{2} / 4) + 1 / 2 \times 2 m_{1} (\sqrt{5} / 4 v_{0})^{2} - 1 / 2 m_{1} v_{0}^{2}$ ,
or $Δ k = \frac{m_{1} v_{0}^{2}}{16} = \frac{m_{1} v_{0}^{2}}{10 + 6}$

Suggest Corrections

thumbs-up

0

Similar questions

m_{1}

with initial speed

v_{0}

in the positive

x

-direction collides with a mass

m_{2} = 2 m_{1}

which is initially at rest at the origin, as shown in figure. After the collision

m_{1}

moves off with speed

v_{1} = v_{0} / 2

in the negative

y

- direction, and

m_{2}

moves off with speed

v_{2}

θ

tan θ

v_{2}

v_{0}

m_{1}

with initial speed

v_{0}

in the positive

x

-direction collides with a mass

m_{2} = 2 m_{1}

which is initially at rest at the origin, as shown in figure. After the collision

m_{1}

moves off with speed

v_{1} = v_{0} / 2

in the negative

y

- direction, and

m_{2}

moves off with speed

v_{2}

θ

. Find the velocity (magnitude and direction) of the centre of mass after the collision :

m_{1}

with initial speed

v_{0}

in the positive

x

-direction collides with a mass

m_{2} = 2 m_{1}

which is initially at rest at the origin, as shown in figure. After the collision

m_{1}

moves off with speed

v_{1} = v_{0} / 2

in the negative

y

- direction, and

m_{2}

moves off with speed

v_{2}

θ

. Write down the

x

y

- components of the equation of conservation of momentum for the collision.

Q. A particle of mass

m_{1}

v_{1}

(in the +x direction) collides with another particle of mass

m_{2}

is at rest before the collision occurs, thus

v_{2} = 0

. After the collision, the particles have velocities

v_{1}^{'}

v_{2}^{'}

in the x-y plane in the directions of

θ_{1}

θ_{2}

with the x-axis (see figure). There re two external forces. Express all out answers in terms of

m_{1}, m_{2}, v_{1}, θ_{1}

θ_{2}

(a) What is the total momentum before the collision (direction and magnitude)
(b) What is the total momentum after the collision (direction and magnitude)
(c) What is the total kinetic energy before the collision?
(d) What is the ratio of the speed

\frac{v_{2}^{'}}{v_{1}^{'}}

(e) What is the magnitude (speed) of

v_{1}^{'}

m_{1} = m_{2}

?

Q. An object of mass

m_{1}

collides elastically with another object of mass

m_{2}

, which is at rest. After the collision, the objects move with equal speeds in opposite directions. The ratio of the masses,

m_{2} : m_{1}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Basketball

PHYSICS

Watch in App

explore_more_icon

Explore more

Definition of a Projectile

Standard XII Physics

Join BYJU'S Learning Program

CrossIcon