Engineering Mechanics Test 1

Concept:

x = 6t² – t³

\(V = \frac{{dx}}{{dt}} = 12t - 3{t^2}\)

 Velocity will be maximum, when

\(\frac{{dv}}{{dt}} = \frac{{d_x^2}}{{d{t^2}}} = 0\)

∴ 12 – 6t = 0

∴ t = 2 seconds

\(\therefore V = \frac{{dx}}{{dt}} = 12t - 3{t^2}\) 

V_max = 12(2) – 3(2)²

∴ V_max = 12 m/s

Concept:

Angular momentum of the system about its center is conserved.

Calculation:

Given:

R = 1m, M = 0.6 kg, m₁ = m₂ = 65g = 0.065 kg

r₁ = 0.4 m, r₂ = 0.6 m, v₁ = v₂ = 25 m/s

Initially, whole system is at rest, therefore

Initial angular momentum (L_i) = 0

After fixing of guns, let system start rotating with angular velocity ‘ω’.

Then,

Final angular momentum (L_f)

L_f = m₁ u₁ r₁ + m₂ u₂ r₂ + I_ω

\({{\rm{L}}_{\rm{f}}}{\rm{\;}} = {m_1}{v_1}\left( {{r_1} + {r_2}} \right) + \frac{1}{2}M{R^2}.\omega \)

\({L_f} = 0.065 \times 25 \times \left( {0.4 + 0.6} \right) + \frac{1}{2} \times 0.6 \times {1^2} \cdot \omega \)

L_f = 1.625 + 0.3 ω

Conservation of angular momentum,

L_i = L_f

0 = 1.625 + 0.3 ω

Concept:

Conservation of momentum:

20 m = m v₁ + m v₂ ; v₁, v₂ → velocity of smooth and rough after impact

\(e = \frac{{velocity\;of\;separation}}{{velocity\;of\;approach}}\)

\(e = \frac{{{v_2} - {v_1}}}{{20}}\)

Perfectly elastic, e = 1

Calculation:

e = 1 ⇒ v₂ – v₁ = 20     

∴ v₂ = v₁ + 20

or  v₁ = v₂ – 20

Also, v₁ + v₂ = 20

∴ v₂ = 20 m/s, v₁ = 0 m/s

Friction retardation on rough block, a = μg

∴ Distance travelled = v² – u² = 2as

⇒ - 400 = - 2μgs

s = 200/μg

∴ s = 101.936 m

Note:

When perfect elastic impact occurs between two identical mass out of which one is at rest, then velocities are exchanged.

Concept:

If the person is sitting in train and observing train velocity, he will observe the train is at rest as the relative velocity of train w.r.t person is zero.

V_TP = V_T – V_P = V_T – V_T = 0

If the person is outside the train and stationery, he will observe the actual speed of train.

V_TP = V_T – V_P = V_T – 0 = V_T

If the person is sitting in another vehicle and observing the train, the observed train velocity will be different.

V_TP = V_T – V_P ⇒ V_T = V_TP + V_P

Hence, the vectors V_T, V_P and V_TP constitute a triangle.

Calculation:

Given u_T = 54 kmph = 15 m/s; u_B = u_P = 0; a_T = 0.2 m/s²; a_P = 0.4 m/s²;

At t = 25 sec,

V_T = u_T + a_T t = 15 + 0.2 × 25 = 20 m/s;

V_P = u_P + a_P t = 0 + 0.4 × 25 = 10 m/s;

Now both the velocities’ directions are different, let’s solve in each direction separately,

We will assume, East direction as positive x – axis and North direction as positive Y – axis.

V_TX = 20 × cos 30° = 17.32 m/s;

V_TY = 20 × sin 30° = 10 m/s;

V_PX = 10 × cos 0° = 10 m/s

V_PY = 10 × sin 0° = 0 m/s;

V_TPX = V_TX – V_PX = 7.32 m/s;

V_TPY = V_TY – V_PY = 10 – 0 = 10 m/s;

Since Both the east component and north component are positive, the direction of Relative velocity vector lies in 1^st quadrant,

∴ Passenger observes the train moving in north east direction (Option 2)

For angle,

\(\tan \theta = \frac{{{V_{TPY}}}}{{{V_{TPX}}}} = \frac{{10}}{{7.32}} = 1.366\) 

⇒ θ = 53.8°

As the angle is measured from East direction, so θ = 53.8° (Option 3)

This problem can also be solved using Velocity triangle approach.

Mistake points:

The angle calculated (θ) is the angle with East direction (+ve x-axis)

If the angle is measured from north direction, then the answer will be 90 - θ degrees.

Concept:

Impulse \((I) = \mathop \smallint \limits_0^t F.dt\)

Calculation:

Given:

F = 600 - 2 × 10⁵ t

Now,

When F = 0

600 – 2 × 10⁵ t = 0

⇒ t = 3 × 10^-3 s

Now,

\(I = \mathop \smallint \limits_0^t F.dt = \mathop \smallint \limits_0^t \left( {600 - 2 × {{10}^5}t} \right)dt = 600t - {10^5}{t^2}\)

I = 600 (3 × 10^-3) – 10⁵ (3 × 10^-3)²

∴ I = 1.8 - 0.9

∴ I = 0.9 N.s

Concept:

It is assumed hitting the ball with a bat as a 1D collision. During a collision, the momentum is conserved.

Momentum lost by bat = Momentum gained by bat

M_bat u_bat - M_bat V_bat = M_ball V_ball - M_ball u_ball

Impulse on the ball will be change in momentum

I = d(mv)

Average force exerted by bat on the ball will be

F = I/Δt

Calculation:

Given:

M_ball = 160 gms = 0.16 kg; M_bat = 1.5 kg; u_ball = 90 kmph = 25 m/s;

V_ball = 144 kmph = 40 m/s; u_bat = 54 kmph = 15 m/s; Δt = 3 ms = 0.003 s

Conservation of momentum:

1.5 × 15 + 0.16 × 25 = 0.16 × 40 + 1.5 × V_bat

⇒ V_bat = 13.4 m/s = 48.24 kmph (Option 4)

Impulse = change in momentum = M_ball (u_ball – (V_ball))

⇒ I = 0.16 (25 – (- 40)) = 10.4 kg m/s (Option 2)

Average force exerted is given by

F_avg = I/Δt = 10.4/0.003 = 3.46 kN (Option3)

Mistake point:

While calculating impulse, directions of velocities are also needed to be considered because directions play a key role in kinematic analysis.

If direction not considered, I = 0.16 × 15 = 2.4 kg.m/s

Concept:

\({V_y} = \frac{{dy}}{{dt}}\)

\({a_y} = \frac{{{d^2}y}}{{dt}}\)

Calculation:

Given

y = x³-3x+150

From, \(\overrightarrow {{V_0}} = 3\hat i - 14\hat j\)

V_x = 3 m/s (Constant)

\({V_y} = \frac{{dy}}{{dt}} = 3{x^2}{V_x} - 3{V_x}\)

\({V_y} = 3{x^2}\left( 3 \right) - 3\left( 3 \right)\)

\({V_y} = 9{x^2} - 9\)      (1)

\({a_y} = 18x\)      (2)

Substitute x = 5 in (1) and (2)

V_y = 216 m/s

a_y = 90 m/s²