In a lab set up, two frictionless carts are placed on a horizontal surface. Cart A has a mass of 500 g and cart B’s mass is 1000 g. Between them is place an ideal spring (very light) and they are squeezed together carefully, thereby compressing the sting by 5.50 cm. Both carts are then released and B’s recoil speed is measured to be 0.55 m/s. A- Will cart’s A speed be:
a- greater than B’s speed
b- less than B’s speed
c- the same as B’s speed

B- Determine A’s recoil speed to see if your conjecture in the first part was correct.

C- Determine the constant of the spring k.

Question

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Respuesta :

nuuk nuuk · Answer 1 · 2019-09-30T12:28:48+02:00

Answer:[tex]v_a=1.1 m/s[/tex]

k=17.32 N/m

Explanation:

Given

mass of cart A =500 gm

mass of cart B =1000 gm

Spring is compressed by a distance of 5.5 cm

B's recoil speed is 0.55 m/s

Conserving momentum

[tex]0+0=0.5\times v_a-1\times 0.55[/tex]

[tex]v_a=1.1 m/s[/tex]

Thus [tex]v_a[/tex] recoil speed is greater than B

Conserving Energy to get value of K

[tex]\frac{m_av^2_a}{2}+\frac{m_bv^2_b}{2}=\frac{kx^2}{2}[/tex]

[tex]\frac{0.5\times 1.1^2}{2}+\frac{1\times 0.55^2}{2}=\frac{k(0.055)^2}{2}[/tex]

k=17.32 N/m