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A 0.400 kg bead slides on a straight frictionless wire and moves with a velocity of 3.50 cm/s to the right, as shown below. The bead collides elastically with a larger 0.600 kg bead that is initially at rest. After the collision, the smaller bead moves to the left with a velocity of 0.70 cm/s.
What is the total kinetic energy of the system of beads after the collision?
A 1.40 × 10−4 J
B 2.45 × 10−4 J
C 4.70 × 10−4 J
D 4.90 × 10−4 J

Respuesta :

Answer:

B

Explanation:

KE = 1/2 m v^2 = 1/2 (0.400) (3.50^2)

= 2.45 J

snehashish65

The total kinetic energy of the system of beads after the collision is [tex]2.45 \times 10^{-4} \;\rm J[/tex]. Hence, option (B) is correct.

Given data:

The mass of bead is, m = 0.400 kg.

The velocity of frictionless wire is, [tex]u = 3.50 \;\rm cm/s = 3.50 \times 10^{-2}\;\rm m/s[/tex]

The mass of larger bead is, m' = 0.600 kg.

The velocity of larger bead is, u' = 0. (Since, it is at rest)

Velocity of smaller bead after the collision is, [tex]v = 0.70 \;\rm cm/s = 0.70 \times 10^{-2} \;\rm m/s.[/tex]

Clearly, we have a hint that the smaller bead has its motion towards left after the collision. This also means that the entire system of beads will move with same speed after the collision, in either direction.

So considering the above fact, the total kinetic energy of the system of beads after the collision is given as,

[tex]KE = \dfrac{1}{2}(m+m')v^{2}[/tex]

Solving as,

[tex]KE = \dfrac{1}{2}(0.400+0.600) \times (0.70 \times 10^{-2})^{2}\\\\KE = 2.45 \times 10^{-4} \;\rm J[/tex]

Thus, we can conclude that the total kinetic energy of the system of beads after the collision is [tex]2.45 \times 10^{-4} \;\rm J[/tex]. Hence, option (B) is correct.

Learn more about the kinetic energy here:

https://brainly.com/question/12669551

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