Engineering Mechanics Test 3

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Engineering Mechanics Test 3

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Weekly Quiz Competition

Question 1
1 / -0

A bullet of mass 20 gm is fixed from a gun of mass 15 kg. If the speed of bullet is 650 m/s, then the recoil velocity of gun is

A
0.5 m/s

B
0.86 m/s

C
1.25 m/s

D
1.5 m/s

Solution

Concept:
We have to use law of conservation of momentum.
Calculation:
Given:
m = 0.02 kg, v₁ = 650 m/s, M = 15 kg
Now,
mv₁ = Mv₂
(0.02) (650) = (15) v₂
∴ V₂ = 0.866 m/s
Question 2
1 / -0

A varying force F = (3t² + 2t + 4) N, (t in sec) acts on a mass of 3 kg. Calculate the velocity if the force acts on mass from t = 1s to t = 5s and the initial velocity is 5 m/s

A
49.67 m/s

B
59.67 m/s

C
39.67 m/s

D
73.4 m/s

Solution

Concept:
Impulse = change in momentum
\(\mathop \smallint \limits_{{t_1}}^{{t_2}} Fdt = \mathop \smallint \limits_u^v dP = m\left( {v - u} \right)\)
Calculation:
\(\mathop \smallint \limits_{{t_1}}^{{t_2}} \left( {3{t^2} + 2t + 4} \right)dt = 3\left( {v - s} \right)\)
\(\therefore \left( {{t^3} + {t^2} + 4t} \right)_1^s = 3\left( {v - s} \right)\)
\(v = \frac{1}{3}\left( {12s + 2s + 20 - 1 - 1 - 4} \right) + s\)
∴ v = 59.67 m/s
Question 3
1 / -0

A horizontal disc is rotating about a vertical axis passing through center with angular velocity of 6π rad/s. A small piece of wax of mass 20 gram falls vertically on disc and sticks at a distance of 10 cm from axis. If angular velocity gets reduced to 5π rad/s, Moment of inertia (gm-m²) of disc is____

A
1-1

B
14-11

C
17-16

D
19-12

Solution

Concept:
Conservation of angular momentum,
L₁ = L₂
∴ I₁ω₂ = I₂ω₂
Calculation:
I₂ = I₁ + mr²
I₂ = I₁ + 20 × 10^-3 × (0.1)²
I₂ = I₁ + 2× 10^-4
I₁ × 6π = (I₁ + 2 × 10^-4) × 5π
1.2 I₁ = I₁ + 2 × 10^-4
0.2 I₁ = 2 × 10^-4
I₁ = 1 × 10^-3
∴ I₁ = 1 gm-m²
Question 4
1 / -0

A glass marble drops from a height of 3 metres upon a horizontal floor. If the co-efficient of restitution be 0.9, what is the height to which it rises after impact?

A
2.43

B
4.43

C
0.61

D
122

Solution

Concept:
If the marble drops from the height h₁ above the floor the velocity with which it would hit the floor:
\(u=\sqrt{2gh_1}\)
Let the velocity of rebound be v and let the marble rises to height h₂after the rebound. Then
\(v=\sqrt{2gh_2}\)
\(Coefficient\;of\;restitution = \frac{{velocity\;of\;separation}}{{veloctiy\;of\;approach}}\)
Calculation:
\(0.9 = \frac{v}{u} = \sqrt {\frac{{2g{h_2}}}{{2g{h_1}}}} \)
\(0.9 = \sqrt {\frac{{{h_2}}}{{{h_1}}}}\)
Now,
Squaring on both sides, we get
\(0.81 = \frac{{{h_2}}}{{{h_1}}}\)
∴ h₂ = 3 × 0.81
∴ h2 = 2.43 m