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3 years ago

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A 0.30 kg mass is sliding on a horizontal, frictionless air track with a speed of 4.0 m/s when it instantaneously hits and stick

s to a 2.0 kg mass initially at rest on the track. The spring constant is 100 N/m . The other end of the spring is fixed.a. Determine the following for the 0.30 kg mass immediately before the impact:i. Its linear momentum ii. Its KEb. Determine the following for the combined masses immediately after the impact:i. The linear momentum in. The KEc. Besides the fact that the objects stick together, what other clue lets you know that this was an inelastic collision

1 answer:

Sveta_85 [38]3 years ago

6 0

Answer

given,

mass of block 1, m = 0.3 Kg

speed of block 1, v = 4 m/s

mass of second block,M = 2 Kg

initial speed of block = 0 m/s

spring constant, k = 100 N/m

a) for block 1

linear momentum before collision

P₁ = m v = 0.3 x 4 = 1.2 Kg.m/s

Kinetic energy

$KE_1 = \dfrac{1}{2}mv^2$

$KE_1 = \dfrac{1}{2}\times 0.3\times 4^2$

$KE_1 =2.4\ J$

b) After impact

final velocity calculation

using conservation of momentum

m v = (m + M )v_f

0.3 x 4 = 2.3 x v_f

v_f = 0.522 m/s

Linear momentum

P₂ = (m+M) v_f

P₂ = 1.5 x 0.522

P₂ = 0.783 kg.m/s

Kinetic energy

$KE_2= \dfrac{1}{2}(M+m)v^2$

$KE_2= \dfrac{1}{2}\times 2.3\times 0.522^2$

$KE_2=0.313\ J$

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