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Mitch throws a 100-g lump of clay at a 500-g target, which is at rest on a horizontal surface. After impact, the target, including the attached clay, slides 2.1 m before stopping. If the coefficient of friction is μ = 0.50, find the speed of the clay before impact.

Respuesta :

Answer:

27.22 m/s

Explanation:

Let the speed of clay before impact is u.

the speed of clay and target is v after impact.

use conservation of momentum

momentum before impact  momentum after impact

mass of clay x u = (mass of clay + mass of target) x v

100 x u = (100 + 500) x v

u = 6 v .....(1)

distance, s = 2.1 m

μ = 0.5

final velocity is zero. use third equation of motion

v'² = v² + 2as

0 = v² - 2 x μ x g x s

v² = 2 x 0.5 x 9.8 x 2.1 = 20.58

v = 4.54 m/s

so by equation (1)

u = 6 x 4.54 = 27.22 m/s

thus, the speed of clay before impact is 27.22 m/s.

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