A block of mass 5 kg moving at a speed of 10 m/s on a frictionless surface collideselastically with another block of same mass which is initially at rest. After collision thefirst block moves at an angle of 370to its initialdirection and has a speed 6 m/s. Estimate the speed of the second block after the collision
Question
A block of mass 5 kg moving at a speed of 10 m/s on a frictionless surface collideselastically with another block of same mass which is initially at rest. After collision thefirst block moves at an angle of 370to its initialdirection and has a speed 6 m/s. Estimate the speed of the second block after the collision
Solution
To solve this problem, we need to use the principles of conservation of momentum and kinetic energy, as the collision is elastic.
Step 1: Conservation of Momentum The total momentum before the collision is equal to the total momentum after the collision.
Before the collision, the momentum is only from the first block, as the second block is at rest. The momentum (p) is calculated as the product of mass (m) and velocity (v), so p1_initial = m1v1_initial = 5 kg * 10 m/s = 50 kgm/s.
After the collision, both blocks are moving. The first block's momentum is p1_final = m1v1_final = 5 kg * 6 m/s = 30 kgm/s. The second block's momentum is unknown, which we'll call p2_final = m2*v2_final.
According to the conservation of momentum, p1_initial = p1_final + p2_final. Substituting the known values, we get 50 kgm/s = 30 kgm/s + p2_final. Solving for p2_final, we find that p2_final = 20 kg*m/s.
Step 2: Find the Velocity of the Second Block We know that p2_final = m2v2_final. We can solve for v2_final by dividing both sides by m2: v2_final = p2_final / m2 = 20 kgm/s / 5 kg = 4 m/s.
Therefore, the speed of the second block after the collision is 4 m/s.
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