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Hunter-Best [27]
3 years ago
6

Each croquet ball in a set has a mass of 0.53 kg. The green ball, traveling at 14.4 m/s, strikes the blue ball, which is at rest

. Assuming that the balls slide on a frictionless surface and all collisions are head-on, find the final speed of the blue ball in each of the following situations: a) The green ball stops moving after it strikes the blue ball. Answer in units of m/s.b) The green ball continues moving after the collision at 2.4 m/s in the same direction. Answer in units of m/s. c) The green ball continues moving after the collision at 0.9 m/s in the same direction. Answer in units of m/s.
Physics
1 answer:
faust18 [17]3 years ago
8 0

a) 14.4 m/s

The problem can be solved by using the law of conservation of total momentum; in  fact, the total initial momentum must be equal to the final total momentum:

p_i = p_f

So we have:

m_g u_g + m_b u_b = m_g v_g + m_b v_b (1)

where

m_b = m_g = m = 0.53 kg is the mass of each ball

u_g = 14.4 m/s is the initial velocity of the green ball

u_b = 0 is the initial velocity of the blue ball

v_g=0 is the final velocity of the green ball

v_b is the final velocity of the blue ball

Simplifying the mass in the equation and solving for v_b, we find

v_b = u_g = 14.4 m/s

b) 12.0 m/s

This time, the green ball continues moving after the collision at

v_g = 2.4 m/s

So the equation (1) becomes

u_g = v_g + v_b

And solving for v_b we find

v_b = u_g - v_g = 14.4 m/s-2.4 m/s=12.0 m/s

c) 13.5 m/s

This time, the green ball continues moving after the collision at

v_g = 0.9 m/s

So the equation (1) becomes

u_g = v_g + v_b

And solving for v_b we find

v_b = u_g - v_g = 14.4 m/s-0.9 m/s=13.5 m/s

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3 years ago
An object of height 2.4 cm is placed 29 cm in front of a diverging lens of focal length 19 cm. Behind the diverging lens, and 11
Arada [10]

Answer:

122.735 behind converging lens ; 2.16

Explanation:

Given tgat:

Object distance, u = 29 cm

Image distance, v =

Focal length, f = - 19 (diverging lens)

Mirror formula :

1/u + 1/v = 1/f

1/29 + 1/v = - 1/19

1/v = - 1/19 - 1/29

1/v = −0.087114

v = −11.47916

v = -11.48

Second lens

Object distance :

u = 11.48 + 11 = 22.48 cm

1/v = 1/19 - 1/22.48

1/v = 0.0081475

v = 1 / 0.0081475

v = 122.735 cm

122.735 behind second lens

Magnification, m

m = m1 * m2

m = - v / u

Lens1 :

m1 = -11.48 / 29 = - 0.3958620

m2 = - 122.735 / 22.48 = - 5.4597419

Hence,

- 0.3958620 * - 5.4597419 = 2.16

8 0
3 years ago
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