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A 2 kg block is moving at a speed of 10 m/s and makes a perfectly elastic collision with a second block of mass m, which is initially at rest. After the collision, the 2 kg block bounces straight back at 3 m/s.

a) Calculate the mass of the second block.
b) Determine the direction of motion of the second block.
c) Find the total kinetic energy after the collision.
d) Calculate the final velocity of the second block.

User Chitza
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Final answer:

Mass and velocity after an elastic collision are found using the conservation of momentum, where the mass of the second block can be calculated, and the direction of its motion is determined. Total kinetic energy remains conserved at 100 Joules, and the second block's final velocity can be found using the calculated mass.

Step-by-step explanation:

a) Calculate the mass of the second block: Using the conservation of momentum: m1 * v1_initial + m2 * v2_initial = m1 * v1_final + m2 * v2_final, 2 kg * 10 m/s + m * 0 m/s = 2 kg * (-3 m/s) + m * v2_final, 20 kg*m/s = -6 kg*m/s + m * v2_final
So, m * v2_final = 26 kg*m/s. b) Determine the direction of motion of the second block: Since momentum is conserved and the 2 kg block bounces back, the second block moves in the original direction of the first block (the +x-direction). c) Find the total kinetic energy after the collision: Since the collision is perfectly elastic, the total kinetic energy is conserved. Initial Kinetic Energy = Final Kinetic Energy. Initial KE = (1/2) * m1 * v1_initial^2, Initial KE = (1/2) * 2 kg * (10 m/s)^2 = 100 Joules. The final KE should also be 100 Joules. d) Calculate the final velocity of the second block: We already have m * v2_final = 26 kg*m/s. Once we have the mass of the second block, we can find its velocity by dividing this momentum by the mass of the second block.

User Sbking
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