You visited us 1 times! Enjoying our articles? Unlock Full Access!

Relative Velocity in 2D

A rod AB of m...

Question

A rod AB of mass 3m and length 4a is freely falling in horizontal position and C is a point at a distant a from A. When speed of rod is u, the point C collides with particle of mass m which is moving vertically upward with speed u. If impact between particle and rod is perfectly elastic then

Velocity of particle after impact

\frac{29}{19} u

downward

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Angular velocity of rod immediately after impact

\frac{12 u}{19 a}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Angular velocity of rod immediately after impact

\frac{18 u}{19 a}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Speed of B immediately after impact

\frac{27 u}{19}

downward

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct options are
A Velocity of particle after impact $\frac{29}{19} u$ downward
B Angular velocity of rod immediately after impact $\frac{12 u}{19 a}$
D Speed of B immediately after impact $\frac{27 u}{19}$ downward

Applying conservation of momentum
3mu - mu = $3 m v_{1} + m v_{2}$
$\Rightarrow 3 v_{1} + v_{2} = 2 u$ . . . .(i)
Applying equation for coefficient of restitution,
$\frac{2 ω - v_{1} + v_{2}}{2 u} = 1$
$\Rightarrow 2 ω - v_{1} + v_{2} = 2 u$ . . . . (ii)
Applying conservation of angular momentum about a point behind center of rod,
$m u a = \frac{3 m (4 a)^{2}}{12} ω - m v_{2} a \Rightarrow m u a = 4 m a^{2} ω - m v_{2} a$
$\Rightarrow u = 4 a ω - v_{2}$ ....(iii)
Solving the three equations,
$v_{2} = \frac{29 u}{19}, ω = \frac{12 u}{19 a} a n d v_{1} = \frac{3 u}{19} t h u s, v_{B} = v_{1} + 2 a ω = \frac{27 u}{19}$

Suggest Corrections

Similar questions

Q. A rigid rod of mass

m

and length

L

is shown in figure. A particle

P

of mass

m

moving with a speed

u

, normal to

A B

strikes

A

and sticks to it. Immediately after the impact, the velocity of

B

w.r.t.

C

n / 10

, where

n

is (in

m/s

)
[

C

is the centre of mass of the system (rod+particle)]

Q. A rod of length

l

forming an anglc

θ

with the horizontal strikes a frictionless floor at A with its centre of mass velocity

v_{0}

and no angular velocity. Assuming that the fimpact atA is perfectly elastic. Find the angular velocity of the rod immediately after the impact

A rod AB of length L and mass M is free to move on a frictionless horizontal surface is moving with a velocity v as shown in fig. End B of rod AB strikes end of wall. Assuming elastic impact, angular velocity of rod AB just after impact is

Q. As shown in figure, two particles

A

and

B

of same mass

2 kg

are joined together by a rigid massless rod of length

2 m

. A particle

P

of mass

2 kg

, moving normal to

A B

with a speed

u = 5 m/s

, strikes at end

A

and sticks to it. The centre of mass of the system

^{'} A + B + P^{'}

C

, then which of the following statement(s) holds true?

Q. A rod of mass M and length L falls from rest from a height H making angle

θ = 37^{\circ}

with horizontal as shown in figure. Coefficient of restitution between rod and ground is 0.5 then just after impact with ground.

Join BYJU'S Learning Program