2) In coming to a stop, a car leaves skid marks on a road that are 40 m long.

A 0.150-kg glider is moving to the right with a speed of 0.80 m/s on a frictionless, horizontal air track. The glider has a head

victus00 [196]

Answer:

The final speed of 0.150-kg glider after collision is 3.2 m/s to the left

The final speed of 0.300-kg glider after collision is 0.20 m/s to the left

Explanation:

Given;

mass of glider moving to the right, m₁ = 0.150-kg

mass of glider moving to the left, m₂ = 0.300-kg

initial speed of glider moving to the right, u₁ = 0.80 m/s

initial speed of glider moving to the left, u₂ = 2.20 m/s

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(0.15 x 0.8) + (0.3 x - 2.2) = 0.15v₁ + 0.3v₂

-0.54 = 0.15v₁ + 0.3v₂

0.15v₁ + 0.3v₂ = - 0.54 -----------equation (i)

Again, their relative velocity after collision is given as;

u₁ - u₂ = v₂ - v₁

0.8 - (-2.2) = v₂ - v₁

3 = v₂ - v₁

v₂ = v₁ + 3 ------------equation (ii)

Substitute v₂ in equation (i)

0.15v₁ + 0.3v₂ = - 0.54

0.15v₁ + 0.3(v₁ + 3 ) = - 0.54

0.15v₁ + 0.3v₁ + 0.9 = - 0.54

0.45v₁ = - 0.54 - 0.9

0.45v₁ = -1.44

v₁ = -1.44 /0.45

v₁ = - 3.2 m/s

Thus, the final speed of 0.150-kg glider after collision is 3.2 m/s to the left

From equation (ii), v₂ = v₁ + 3

v₂ = -3.2 + 3

v₂ = - 0.20 m/s

Therefore, the final speed of 0.300-kg glider after collision is 0.20 m/s to the left

3 0

4 years ago

What does the term calorie refer to

OLga [1]

Hey there,
<span>It means- the amount of energy required to raise the temperature of a liter of water one degree centigrade at sea level
</span>
Hope this helps :))

~Top

6 0

4 years ago

Why do sunspots appear darker than their surroundings?

andre [41]

They are cooler than their surroundings<span> and they're cooler because of the intense magnetic fields in them</span>

7 0

3 years ago

Read 2 more answers

Why do racing cars have wider wheels?

tamaranim1 [39]

Answer:

So they can last longer and have more grip than normal on-road cars. They need that in order for them to run well

Explanation:

3 0

3 years ago

Read 2 more answers

A mass attached to a 66.4 cm long string starts from rest and is rotated 36.6 rev in1 min before reaching a final angular speed.

andrey2020 [161]

Answer:

0.128 rad/s², 7.66 rad/s

Explanation:

length, l = 66.4 cm

initial angular speed, ωo = 0 rad/s

Let ω be the final angular speed.

Let α be the angular acceleration.

number of revolution, n = 36.6

time, t = 1 min = 60 second

Angle turned, θ = 2πn = 2 x 3.14 x 36.6 = 229.85 rad

Use second equation of motion for angular motion

$\theta =\omega _{0}t+\frac{1}{2}\alpha t^{2}$

229.85 = 0 + 0.5 x α x 60 x 60

α = 0.128 rad/s²

Use first equation of motion

ω = ωo + αt

ω = 0 + 0.128 x 60

ω = 7.66 rad/s

5 0

4 years ago