Engineering Mechanics Test 3

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

Concept:

Impulse = change in momentum

$\underset{t_1}{\overset{t_2}{\int }}Fdt=\underset{u}{\overset{v}{\int }}dP=m\left(v-u\right)$

Calculation:

$\underset{t_1}{\overset{t_2}{\int }}\left(3t^2+2t+4\right)dt=3\left(v-s\right)$

$\therefore {\left(t^3+t^2+4t\right)}_1^s=3\left(v-s\right)$

$v=\frac{1}{3}\left(12s+2s+20-1-1-4\right)+s$

∴ v = 59.67 m/s

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

Concept:

If the marble drops from the height h₁ above the floor the velocity with which it would hit the floor:

$u=\sqrt{2g{h}_1}$

Let the velocity of rebound be v and let the marble rises to height h₂ after the rebound. Then

$v=\sqrt{2g{h}_2}$

$Coefficientofrestitution=\frac{velocityofseparation}{veloctiyofapproach}$

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