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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.

9. What is the large bead’s velocity after the collision?



Answer :

MathPhys

Answer:

2.8 cm/s to the right

Explanation:

The momentum of an object (p) is defined as the product of its mass (m) and its velocity (v), which is the magnitude and direction of its speed. When there are no external forces, the total momentum of a system is conserved before and after the collision.

p₁ = p₂

m₁ u₁ + m₂ u₂ = m v₁ + m₂ v₂

Since the larger bead is initially at rest, u₂ = 0. The equation simplifies to:

m₁ u₁ = m v₁ + m₂ v₂

Plug in values:

(0.400 kg) (3.50 cm/s) = (0.400 kg) (-0.70 cm/s) + (0.600 kg) v₂

Solving:

v₂ = 2.8 cm/s

The large bead moves at 2.8 cm/s to the right.

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