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4. A 25,000kg asteroid is flying towards the Earth at a speed of 1500.0m/s. How much Silly Putty would it take to stop the asteroid if we launched it at a velocity of -100.0m/s? Assume the Silly Putty sticks to the asteroid.​

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To solve this problem, we can use the principle of conservation of momentum, which states that the total momentum of a system remains constant if there are no external forces acting on it. We can set the momentum of the asteroid before the collision equal to the momentum of the asteroid-Silly Putty system after the collision.

The momentum of the asteroid before the collision is:

P_before = m_ast * v_ast

where m_ast is the mass of the asteroid and v_ast is its velocity.

The momentum of the asteroid-Silly Putty system after the collision is:

P_after = (m_ast + m_sp) * v_final

where m_sp is the mass of the Silly Putty, v_final is the velocity of the asteroid-Silly Putty system after the collision, which we assume to be zero.

We can equate these two expressions for momentum and solve for the mass of the Silly Putty:

m_sp = (m_ast * v_ast) / (-v_final)

Substituting the given values:

m_ast = 25,000 kg

v_ast = 1500.0 m/s

v_final = -100.0 m/s

m_sp = (25,000 kg * 1500.0 m/s) / (-(-100.0 m/s))

m_sp = 375,000 kg m/s / 100.0 m/s

m_sp = 3750 kg

Therefore, we would need 3750 kg of Silly Putty to stop the asteroid if we launched it at a velocity of -100.0 m/s.