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Fantom [35]
3 years ago
14

A 5 kg bowling ball with a velocity of +10 m/s collides with a stationary 2 kg bowling pin. If the ball's final velocity is +8 m

/s, what is the pin's final velocity?
a 5 m/s
b 2.5 m/s
c 10 m/s
d 5.2 m/s
Physics
1 answer:
Whitepunk [10]3 years ago
3 0

Answer:

<h2>The pin's final velocity is 5m/s</h2>

Explanation:

Step one:

given data

mass of ball m1=5kg

initial velocity of ball u1=10m/s

mass of pin m2=2kg

initial velocity of pin u2= 0m/s

final velocity of ball v2=8m/s

final velocity of pin v2=?

Step two:

The expression for elastic collision is given as

m1u1+m2u2=m1v1+m2v2

substituting we have

5*10+2*0=5*8+2*v2

50+0=40+2v2

50-40=2v2

10=2v2

divide both sides by 2

v2=10/2

v2=5m/s

The pin's final velocity is 5m/s

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6 0
3 years ago
*A car is going through a dip in the road whose curvature approximates a circle of radius 150m. At what velocity will the occupa
Valentin [98]

Answer:

v= 14.85 m/s

Explanation:

  • When at the bottom of the dip, the only force that keeps the car in the circular trajectory, is the centripetal force.
  • This force is not a new force, is just the net force aiming to the center of the circle.
  • In this case, is just the difference between the normal force (always perpendicular to the surface, pointing upward) and the force that gravity exerts on the car (which is known as the weight), pointing downward.
  • So, we can write the following expression:

       F_{cent} = F_{n} - F_{g}  (1)

  • It can be showed that the centripetal force is related to the speed by the following expression:
  • F_{cent} = m*\frac{v^{2}}{r} (2)
  • The normal force, it is called the apparent weight, because it would be the weight as measured by a scale.
  • Replacing (2) in (1), and solving for Fn, we get:

       F_{n} = m*\frac{v^{2} }{r} + m*g (3)

  • Now, we need to find the value of v that makes Fn, exactly 15% more than the weight m*g, so we can write the following equation:

      F_{n} = 1.15*F_{g} = m*\frac{v^{2}}{r} +F_{g}  (4)

  • Replacing Fg by its value, simplifying, and solving for v, we get:

       v = \sqrt{0.15*g*r} = \sqrt{9.8 m/s2*0.15*150m} = 14.85 m/s (5)

3 0
2 years ago
how fast is a ball going when it hits the ground after being dropped from a height of 16m the acceleration of gravity is 9.8
Aneli [31]

Answer:

17.7 m/s

Explanation:

Given:

y₀ = 0 m

y = 16 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: v

v² = v₀² + 2a (y − y₀)

v² = (0 m/s)² + 2 (9.8 m/s²) (16 m − 0 m)

v = 17.7 m/s

The ball is moving at a speed of 17.7 m/s when it hits the ground.

8 0
3 years ago
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