A 88-kg airplane wheel is moving east at 2.41 m/s and strikes a stationary 95-kg chock. Immediately after the head-on collision, the choke is moving east at 5.19 m/s. Determine the post collision velocity of the airplane wheel?
Answers
Answer:
ak47
Explanation:
Given: A 88 kg airplane wheel is moving east at 2.41 m/s and strikes a stationary 95 kg chock. Immediately after head on collision, the choke is moving east at 5.19 m/s
To find: The post collision velocity of the airplane wheel
Explanation: When head on collision occurs, the momentum of the system does not change, it is always balanced.
Momentum of 88 kg airplane before collision= 88* 2.41 m/s = 212.08
Momentum of chock before collision= 0
Momentum of 95 kg chock after collision= 95 * 5.19 m/s = 493.05
Let p be the momentum of the airplane after collision.
Momentum before collision= Momentum after collision
=> 212.08 +0 = 493.05+ p
=> p = 212.08-493.05
= - 280.97
Therefore, minus sign signifies airplane is moving with some velocity in the direction opposite to its previous direction, that is in the west direction.
Velocity= Momentum/ mass of airplane
= -280.97 / 88
= -3.19 m/s
Therefore, the velocity of the airplane after collision is 3.19 m/s in west direction.