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Math  /  Algebra

QuestionA 2.00−kg2.00-\mathrm{kg} mass is given an initial velocity by a spring-loaded mechanism on a frictionless surface. This mass impacts a 3.00-kg mass sitting on the edge of the table. The 2.00-kg mass ends up moving backwards at 1.31 m/s\mathrm{m} / \mathrm{s}. How far from the base of the 1.27-meter-high table does the 3.00−kg3.00-\mathrm{kg} mass land? The spring constant is 404 Newtons/meter and the spring is initially compressed 0.164 meters.

Studdy Solution

STEP 1

1. The collision between the two masses is perfectly elastic.
2. The surface is frictionless, so no energy is lost to friction.
3. The only forces acting on the masses after the collision are gravitational.
4. The spring follows Hooke's Law, providing potential energy to the 2.00-kg mass.
5. The 3.00-kg mass falls freely under gravity after the collision.

STEP 2

1. Calculate the initial velocity of the 2.00-kg mass using energy conservation.
2. Use conservation of momentum to find the velocity of the 3.00-kg mass after the collision.
3. Calculate the time of flight for the 3.00-kg mass as it falls from the table.
4. Determine the horizontal distance traveled by the 3.00-kg mass.

STEP 3

Calculate the initial velocity of the 2.00-kg mass using the potential energy stored in the spring. The potential energy PE PE is given by:
PE=12kx2 PE = \frac{1}{2} k x^2
where k=404 N/m k = 404 \, \text{N/m} is the spring constant and x=0.164 m x = 0.164 \, \text{m} is the compression. The kinetic energy KE KE of the 2.00-kg mass is:
KE=12mv2 KE = \frac{1}{2} m v^2
Set the potential energy equal to the kinetic energy to solve for the initial velocity v v :
12kx2=12mv2 \frac{1}{2} k x^2 = \frac{1}{2} m v^2
v=kx2m v = \sqrt{\frac{k x^2}{m}}
v=404×(0.164)22.00 v = \sqrt{\frac{404 \times (0.164)^2}{2.00}}
v≈2.33 m/s v \approx 2.33 \, \text{m/s}

STEP 4

Use conservation of momentum to find the velocity of the 3.00-kg mass after the collision. The initial momentum pi p_i is:
pi=m1v1+m2v2=2.00×2.33+3.00×0 p_i = m_1 v_1 + m_2 v_2 = 2.00 \times 2.33 + 3.00 \times 0
The final momentum pf p_f is:
pf=m1v1′+m2v2′=2.00×(−1.31)+3.00×v2′ p_f = m_1 v_1' + m_2 v_2' = 2.00 \times (-1.31) + 3.00 \times v_2'
Set initial momentum equal to final momentum:
2.00×2.33=2.00×(−1.31)+3.00×v2′ 2.00 \times 2.33 = 2.00 \times (-1.31) + 3.00 \times v_2'
Solve for v2′ v_2' :
4.66=−2.62+3.00×v2′ 4.66 = -2.62 + 3.00 \times v_2'
3.00×v2′=7.28 3.00 \times v_2' = 7.28
v2′=7.283.00 v_2' = \frac{7.28}{3.00}
v2′≈2.43 m/s v_2' \approx 2.43 \, \text{m/s}

STEP 5

Calculate the time of flight for the 3.00-kg mass as it falls from the table. Use the equation for free fall:
h=12gt2 h = \frac{1}{2} g t^2
where h=1.27 m h = 1.27 \, \text{m} and g=9.81 m/s2 g = 9.81 \, \text{m/s}^2 . Solve for t t :
1.27=12×9.81×t2 1.27 = \frac{1}{2} \times 9.81 \times t^2
t2=2×1.279.81 t^2 = \frac{2 \times 1.27}{9.81}
t2≈0.2588 t^2 \approx 0.2588
t≈0.2588 t \approx \sqrt{0.2588}
t≈0.5087 s t \approx 0.5087 \, \text{s}

STEP 6

Determine the horizontal distance traveled by the 3.00-kg mass. Use the horizontal velocity v2′ v_2' and the time of flight t t :
distance=v2′×t \text{distance} = v_2' \times t
distance=2.43×0.5087 \text{distance} = 2.43 \times 0.5087
distance≈1.24 m \text{distance} \approx 1.24 \, \text{m}
The 3.00-kg mass lands approximately 1.24 m 1.24 \, \text{m} from the base of the table.

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