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bija089 [108]
3 years ago
12

In a plate glass factory, sheets of glass move along a conveyor belt at a speed of 15.0 cm/s. An automatic cutting tool descends

at preset intervals to cut the glass to size. Since the assembly belt must keep moving at constant speed, the cutter is set to cut at an angle to compensate for the motion of the glass. If the glass is 72.0 cm wide and the cutter moves across the width at a speed of 24.0 cm/s, at what angle should the cutter be set
Physics
1 answer:
Leni [432]3 years ago
3 0

Answer:

angle at which the cutter should be set = 32°

Explanation:

We are given;

Speed of conveyor belt; v1 = 15 cm/s

Width;w = 72 cm

Speed of the cutter that moves across the width; v2 = 24 cm/s

Let's calculate time;

Since the width and the rate at which the cutter works per seconds is given, thus;

Time = w/v2

Time = 72/24 = 3 seconds

Now, we know that, velocity = distance/time

Thus, distance = velocity x time

So, distance covered;d = 15 x 3 = 45 cm

Now, for us to calculate the angle the cutter should be set, we would treat it as a triangle.

When the cutter cuts it, it covers a distance d which is perpendicular to the width.

Thus, to calculate the angle made with the width, we'll use trigonometric ratio.

Thus, d/w = tan θ where θ is the angle at which the cutter should be set.

So, 45/72 = tan θ

tan θ = 0.625

θ = tan^(-1)0.625

θ = 32°

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A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.
arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

  • Mass of the first puck = m_1\ =\ 5\ kg
  • Mass of the second puck = m_2\ =\ 3\ kg
  • initial velocity of the first puck = u_1\ =\ 3\ m/s.
  • Initial velocity of the second puck = u_2\ =\ -1.5\ m/s.

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = v_2\ =\ 2.31\ m/s.

Let v_1 be the final velocity of first puck after the collision.

From the conservation of linear momentum,

m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = 25\times 10^{-3}\ sec

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

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