A 0.329-kg puck, initially at rest on a frictionless horizontal surface, is struck by a 0.222-kg puck that is intitially moving along the x-axis with a velocity of 2.24 m/s. After the collision, the 0.222-kg puck has a speed of 1.24 m/s at an angle of 24 degrees to the positive x-axis. (a) Determine the velocity of the 0.329-kg puck after the collision. (b) Find the fraction of the kinetic energy lost in the collision.
2007-11-25 15:43:52 · 2 answers · asked by grouchy187 2 in Science & Mathematics ➔ Physics
Momentum is conserved in both dimensions.
Initial x momentum: (.222 kg)(2.24 m/s)
Initial y momentum: zero
Final x momentum: (.222 kg)(1.24 m/s)(cos 24) + (.329)(vx)
Final y momentum: (.222 kg)(1.24 m/s)(sin 24) + (.329)(vy)
Initial x momentum = final x momentum
Solve for vx, the x component of the velocity of the second puck.
Initial y momentum = final y momentum
Solve for vy, the y component of the velocity of the second puck.
Initial kinetic energy = 1/2 (.222 kg)(2.24 m/s)^2
Final kinetic energy = 1/2 (.222 kg)(1.24 m/s)^2 + 1/2 (.329 kg)(vx^2 + vy^2)
Compute both, then determine the fraction lost.
2007-11-25 16:09:03 · answer #1 · answered by jgoulden 7 · 0⤊ 0⤋
For elastic collisions I worked out here to keep time answering ?s in basic terms like yours: First discover the cost of the midsection of mass of the two gadgets: Vc = (m1v1 + m2v2)/(m1 + m2)......this could be + or - based on the values of v1 & v2. Then discover u1 & u2, the cost of each mass relative to the CM: u1 = ?V/(a million+ m1/m2)..... the place ?V = l v1-v2 l u2 = ?V/(a million+ m2/m1) In an elastic collision those would be a similar in the previous and after the collision, so in basic terms upload or subtract them from Vc to discover v1f & v2f
2016-10-02 04:25:08 · answer #2 · answered by ? 4 · 0⤊ 0⤋