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Yuki888 [10]
3 years ago
15

A block of mass mm starts from rest and slides down from the top of a wedge of height hh and length dd. The surface of the wedge

forms an angle of \thetaθ with respect to the horizontal direction. The force of kinetic friction between the block and the wedge is \vec{f} f ​ . How fast is the block traveling when it reaches the bottom of the wedge?
Physics
1 answer:
lutik1710 [3]3 years ago
6 0

Answer:

v=\sqrt{ 2 (g sin\theta- f/m) d}

Explanation:

Given that

Mass = m

Height = h

Length = d

Angle = θ

Friction force = f

When block slides downward then friction force will act upwards

The force due to gravity along inline plane

Fg= mg sinθ

Lets 'a' is the acceleration

From Newtons law

mg sinθ  - f = m a

a= (g sinθ  - f/m)

Given that initial speed u= o ,lets v is final speed

v²=u²+ 2 a s

v²=2 x (g sinθ  - f/m) x d

v=\sqrt{ 2 (g sin\theta- f/m) d}

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3 years ago
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.
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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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3 years ago
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