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Free_Kalibri [48]
3 years ago
8

After being struck by a bowling ball, a 1.7 kg bowling pin sliding to the right at 3.8 m/s collides head-on with another 1.7 kg

bowling pin initially at rest. Find the final velocity of the second pin in the following situations: a) The first pin moves to the right after the collision at 0.8 m/s. Answer in units of m/s.
Physics
1 answer:
Alchen [17]3 years ago
7 0

Answer:

3 m/s

Explanation:

Parameters given:

Mass of first bowling pin, m = 1.7 kg

Initial velocity of first bowling pin, u = 3.8 m/s

Final velocity of first bowling pin, v = 0.8 m/s

Mass of second bowling pin, M = 1.7 kg

Initial velocity of second bowling pin, U = 0 m/s

Let the final velocity of the second bowling pin be V

Using the principle of conservation of momentum:

Total initial momentum = Total final momentum

m*u + M*U = m*v + M*V

(1.7 * 3.8) + 0 = (1.7 * 0.8) + (1.7 * V)

6.46 = 1.36 + 1.7V

1.7V = 5.1

V = 5.1/1.7 = 3 m/s

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Picture #1:
GPE = (mass) x (gravity) x (height)
GPE = (2 kg) x (9.8 m/s²) x (40 m) = 784 joules

KE = (1/2) (mass) (speed²)
KE = (1/2) (2 kg) (5 m/s)²
KE = (1 kg) (25 m²/s²)  =  25 joules

Picture #2:
KE = (1/2) (mass) (speed²)
KE = (1/2) (2 kg) (10 m/s)²
KE = (1 kg) (100 m²/s²)  =  100 joules

Picture #3:
GPE = (mass) x (gravity) x (height)
GPE = (20 kg) x (9.8 m/s²) x (2 m) = 392 joules

KE = (1/2) (mass) (speed²)
KE = (1/2) (20 kg) (5 m/s)²
KE = (10 kg) (25 m²/s²)  =  250 joules

Picture #4:
GPE = (mass) x (gravity) x (height)
98 joules = (1 kg) x (9.8 m/s²) x (height)
Height = (98 joules) / (1 kg x 9.8 m/s²)
Height = 10 meters

Picture #5:
GPE = (mass) x (gravity) x (height)
39,200 Joules = (mass) x (9.8 m/s²) x (20 m)
Mass = (39,200 joules) / (9.8 m/s² x 20 m)
Mass = 200 kg

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