GATE ME 7. Engineering Mechanics Chapter Engineering Mechanics Test 1 Mock Test Series 2022-23

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Engineering Mechanics Test 1

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Weekly Quiz Competition

Question 1
2 / -0.33

A particle moves in a straight line and its position is defined by equation $x = 6 t^{2} - t^{3}$ where t is expressed in seconds and x in meters. The maximum velocity(m/s) during motion is____

A
12-12

B
124-121

C
127-126

D
129-122

Solution

Concept:
x = 6t² – t³
$V = \frac{d x}{d t} = 12 t - 3 t^{2}$
Velocity will be maximum, when
$\frac{d v}{d t} = \frac{d_{x}^{2}}{d t^{2}} = 0$
∴ 12 – 6t = 0
∴ t = 2 seconds
$∴ V = \frac{d x}{d t} = 12 t - 3 t^{2}$
V_max = 12(2) – 3(2)²
∴ V_max = 12 m/s
Question 2
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A player hits a ball with a baseball but such that it repeats its path and goes back with incoming velocity of 20 m/s. if the mass of ball is 150 gm, then the impulse (magnitude) imparted by the bat is ____ kg.m/s (give magnitude only)

A
6-6

B
64-61

C
67-66

D
69-62

Solution

Concept:
Impulse (I) = change in momentum
Calculation:
Given:
V₁ = 20 m/s, V₂= -20 m/s, M = 150 g = 0.15 kg
Now,
Impulse (I) = change in momentum
I = m (V₂ – V₁)
I = 0.15 (-20 - 20)
∴ I = - 6 kg m/s. (∵ it has been asked about the magnitude only hence the answer will be 6)
Question 3
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Two blocks of equal masses are made. One of them is smooth and the other is rough with coefficient of friction 0.20 with the floor. The smooth block is moving at 20 m/s towards the rough block at rest. If their collision is perfectly elastic, then the distance travelled by the rough block will be ______

A
100-103

B
1004-1031

C
1007-1036

D
1009-1032

Solution

Concept:
Conservation of momentum:
20 m = m v₁ + m v₂ ; v₁, v₂ → velocity of smooth and rough after impact
$e = \frac{v e l o c i t y o f s e p a r a t i o n}{v e l o c i t y o f a p p r o a c h}$
$e = \frac{v_{2} - v_{1}}{20}$
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.
Question 4
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A horizontal circular platform of radius 1m and mass 0.6 kg is free to rotate about its axis. Two massless guns, each carrying a bullet of mass 65 gm are attached at a distance of 0.4 m and 0.6 m from center and in opposite direction. If the guns fire simultaneously and speed of bullets are 25 m/s (each), then the magnitude of rotational speed of platform is

A
2.71 rad/s

B
3.27 rad/s

C
5.41 rad/s

D
5.95 rad/s

Solution

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
$L_{f} = m_{1} v_{1} (r_{1} + r_{2}) + \frac{1}{2} M R^{2} . ω$
$L_{f} = 0.065 \times 25 \times (0.4 + 0.6) + \frac{1}{2} \times 0.6 \times 1^{2} \cdot ω$
L_f = 1.625 + 0.3 ω
Conservation of angular momentum,
L_i = L_f
0 = 1.625 + 0.3 ω
∴ ω = - 5.41 rad/s
Question 5
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A bullet is fired from a gun. The force on the bullet is given by F = 600 – 2 × 10⁵ t where F is in newton and t in seconds. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?

A
9 N-s

B
Zero

C
0.9 N-s

D
1.8 N-s

Solution

Concept:
Impulse $(I) = \int_{0}^{t} F . d t$
Calculation:
Given:
F = 600 - 2 × 10⁵ t
Now,
When F = 0
600 – 2 × 10⁵ t = 0
⇒ t = 3 × 10^-3s
Now,
$I = \int_{0}^{t} F . d t = \int_{0}^{t} (600 - 2 \times 10^{5} t) d t = 600 t - 10^{5} t^{2}$
I = 600 (3 × 10^-3) – 10⁵ (3 × 10^-3)²
∴ I = 1.8 - 0.9
∴ I = 0.9 N.s