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A 115 kg hockey player, Adam, is skating east when he tackles a 133 kg player, Bob, who is skating west at -1.59 m/s. Afterwards, they move together at 0.250 m/s east. What was Adam's velocity before the collision? (Unit = m/s)

Answer :

Answer:

2.38 m/s

Explanation:

Momentum is conserved:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

After the collision, they move at the same speed, so v₁ = v₂ = v.

m₁u₁ + m₂u₂ = (m₁ + m₂) v

Plugging in numbers:

(115 kg) u₁ + (133 kg) (-1.59 m/s) = (115 kg + 133 kg) (0.250 m/s)

u₁ = 2.38 m/s

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Rewritten by : Barada

Final answer:

To find Adam's velocity before the collision, we can use the law of conservation of momentum. By setting up an equation using the mass and velocity of Adam and Bob before the collision, we can solve for Adam's velocity. The velocity is found to be 1.976 m/s.

Explanation:

To solve this problem, we can use the law of conservation of momentum, which states that the total momentum of a system before a collision is equal to the total momentum after the collision.

Let's assume Adam's velocity before the collision is v.

Using the law of conservation of momentum:

(mass of Adam) * (velocity of Adam before the collision) + (mass of Bob) * (velocity of Bob before the collision) = (mass of Adam + mass of Bob) * (velocity after the collision)

(115 kg) * (v m/s) + (133 kg) * (-1.59 m/s) = (115 kg + 133 kg) * (0.25 m/s)

115v - 211.47 = 53 * 0.25

115v = 15.75 + 211.47

115v = 227.22

v = 227.22 / 115

v = 1.976 m/s