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A 9.0 kg bowling ball races down the lane at 15 m/s before striking a bowling pin (at rest) with a mass of 0.85 kg. If the 0.85 kg pin bounces backward with a velocity of 45 m/s, what is the velocity of the bowling ball after the collision

Respuesta :

xero099

Answer:

[tex]v = 10.75\,\frac{m}{s}[/tex]

Explanation:

The system ball-pin is modelled by the Principle of Moment Conservation:

[tex](9\,kg)\cdot (15\,\frac{m}{s} ) + (0.85\,kg)\cdot (0\frac{m}{s} ) = (9\,kg)\cdot v + (0.85\,kg)\cdot (45\,\frac{m}{s} )[/tex]

The velocity of the bowling ball after the collision is:

[tex]v = 10.75\,\frac{m}{s}[/tex]

akande212

Answer:

v1 = 10.75m/s

Explanation:

Given m = 9.0kg

u1 = 15m/s, m2 = 0.85kg, u2 = 0m/s bowling pin initially at rest. v1 =? v2 = 45m/s

Where u represents initial velocity and v final velocity for the bodies involved in the collision.

From the principle of conservation of momentum, the sum of momentum before collision is equal the sum of momentum after collision.

m1u1 + m2u2 = m1v1 + m2v2

9×15 + 0.85×0 = 9v1 + 0.85×45

135 = 9v1 +38.25

9v1 = 135 – 38.25

9v1 = 96.75

v1 = 96.75/9 = 10.75m/s