GATE Past Year Questions: Impulse & Momentum | Engineering Mechanics - Civil Engineering (CE) PDF Download

r = t ²
θ = t
x = r cos θ = t²cos t
y = r sin θ = t² sin t

at
t = 2 u = 4 cos 2 – 4 si n 2 = – 5.301
 = 2t sin t + t² cos t
at
t = 2 v = 4 sin 2 + 4 cos 2 = 1.972


= 1/2  x 1 x 5.656² = 15.99

I = 5mr²/2
K.E. = (K.E.) _translation + (K.E.) _Rotation
= 1/2mv² + 1/2 Iω²
where V = 2 R ω
⇒ ω = V/2R
∴ K.E 1/2 mV² + 5mV²/16
K.F = 13mV²/16

Energy stored in the bar = Work done by the stone

= 300 x³/3
= 100 x 0.1³
= 0.1J

Power = dW/dt (rate of doing work or rate of change of K.E. energy of the system) Initial Kinetic energy of the system
= 1/2 mv² + 1/2 mv² = mv²
Final Kinetic energy = 0
Time duration it occurs = t
⇒ Power = mv²/t

Method I:

1/2K x² will be converted into potential energy (taking reference as compressed condition)
1/2K x²=mgh
1/2×981×10³×(0.1)² = 100×9.81h
h = 5m = 5000mm
 

Method II:

When spring will reach its natural length, K.E. of the block of mass 100 kg held on it.
(taking reference at natural length of spring)
1/2k x²−mgx=1/2mV² [x=0.1m]
and this K.E will be converted into mgh.
Total lift of block = h+x
1/2×981×10³×.1−100×9.81×.1 = mgh = 100×9.81×h
h = 4.9 m
⇒ h+x = 4.9+0.1 = 5m

Disc A → I = mR² + mR² = 2mR²
Disc B → I = mR²/2 + mR² = 3/2 mR²

Loss in potential energy = Gain in kinetic energy
Mgh = 1/2 MV²
V =

Coefficient of restitution
= Relative speed after collision/Relative speed before collision = 0

At t = 0, mass is at rest, F = 5 kN for 10^–4 seconds & amplitude = x
F = m x dv x dt

F(t₂ – t₁) = m × (v₂ – v₁)
5 × 10^–3 × 10^–4 = 1 × (v₂ – 0)
0.5 m/s = v₂
 = 1/2kx²

Linear momentum and Kinetic Energy is conserved in elastic collision and in inelastic collision Kinetic Energy (heat, sound etc.) is conserved.

Conservation of linear momentum,
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
⇒ u₂ = 0
3 x 4 = 3 x v₁ + mv₂ ...(i)
Coefficient of restitution , e = 1 for perfectly elastic collision,
v₂ - v₁ = (u₁ - u₂)
⇒ v₂ - v₁ = u₁ = 4
⇒ v₂ - v₁ = 4 ...(ii)
Conservation of energy,

⇒ (24 - 6)2 =
⇒  = 36 ...(iii)
∵ v₁ = v₂ - 4
Putting in equation (i)
⇒ 12 = 3v₂ - 12 + mv₂
⇒ v₂(3 + m ) = 24 ...(iv)
Putting in equation (iii),

⇒ (3 + m)² = 16m
⇒ 9 + m² + 6m – 16m = 0
⇒ m² – 10m + 9 = 0
⇒ m² – 9m – m + 9 = 0
(m – 9)(m – 1) = 0
m = 1, 9 k g

For perfectly plastic impact, Coefficient of restitution
= Relative velocity of separation/Relative velocity of approach = 0

Taking both clay & wheel as system
L_i = L_f (about A)


L_i = mvr = 1×10×1=10
L_f = mv_cmr + Iω
= 20 x rω x r + m²ω/2 = 20ω + 10ω
L_f = 30ω
L_i = Lf
10 = 30ω
ω = 1/3 rad/sec

Let, m = mass of the bullet
v = velocity of bullet
M = mass of the block
u = final velocity of system (block + bullet)

From conservation of linear momentum
mv = (M + m)u
⇒ u = mv/ m + M
For horizontal equilibrium of body
(M + m)g = N
F_r = Frictional force = μN = u(M + m)g
Frictional retardation =  = -μg

(–ve sign shows that acceleration is opposite to motion)
Let vf = final velocity of system = 0
= u² + 2as
u² + 2as = 0
u² – 2mgs = 0
From equation (i)

In a perfectly elastic collision between equal masses of two bodies, velocities exchange on impact.
Velocity just before impect = Velocity immediately after impact.
Before Impact
V_A = 
V_B = 0

After Impact
V_A = 0
V_B =