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bonufazy [111]
3 years ago
10

A 5 kg ball rolling at 1.0 m/s hits a 15 kg ball at rest. The balls stick together after the collision What is the

Physics
1 answer:
Reika [66]3 years ago
5 0

Answer:

0.25m/s

Explanation:

Given parameters

m₁  = 5kg

v₁ = 1.0m/s

m₂ = 15kg

v₂ = 0m/s

Unknown:

velocity after collision = ?

Solution:

Momentum before collision and after collision will be the same. For inelastic collision;

    m₁v₁ + m₂v₂  = v(m₁ + m₂)

Insert parameters and solve for v;

   5 x 1  + 15 x 0 = v (5 + 15 )

          5  = 20v

          v = \frac{5}{20}   = 0.25m/s

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physics A river flows at a speed vr = 5.37 km/hr with respect to the shoreline. A boat needs to go perpendicular to the shorelin
vova2212 [387]

Answer: Vb is the vector  (-5.37m/s,  8.59 m/s), with a module 10.13m/s

then the ratio Vb/Vr = 10.13m/s/5.37m/s = 1.9

Explanation:

We can use the notation (x, y) where the river flows in the x-axis and the pier is on the y-axis.

We have Vr = (5.37m/s, 0m/s)

Now, if the boat wants to move only along the y-axis (perpendicularly to the shore).

The velocity of the boat Vb will be:

Vb = (-c*sin(32). c*cos(32))

Then we should have that:

5.37 m/s - c*sin(32) = 0

c = (5.37/sin(32))m/s = 10.13 m/s

the velocity in the y-axis is:

10.13m/s*cos(32) = 8.59 m/s

So Vb = (-5.37m/s,  8.59 m/s)

the ratio Vb/Vr = 10.13m/s/5.37m/s = 1.9 where i used Vb as the module of the boat's velocity.

7 0
2 years ago
A horizontal 2.00\ m2.00 m long, 5.00\ kg5.00 kg uniform beam that lies along the east-west direction is acted on by two forces.
Sunny_sXe [5.5K]

Answer: 240\ rad/s^2

Explanation:

Given

Length of beam l=2\ m

mass of beam m=5\ kg

Two forces of equal intensity acted in the opposite direction, therefore, they create a torque of magnitude

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I=\dfrac{ml^2}{12}=\dfrac{5\times 2^2}{12}\\\\I=\dfrac{5}{3}\ kg.m^2

Torque is the product of moment of inertia and angular acceleration

\Rightarrow \tau=I\alpha\\\\\Rightarrow 400=\dfrac{5}{3}\times \alpha\\\\\Rightarrow \alpha =240\ rad/s^2

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