Problem 7.25 Part A What is be the speed of the 0.444-kg ball after the collisio

Question

Problem 7.25 Part A What is be the speed of the 0.444-kg ball after the collision? Express your answer to three significant figures and include the appropriate units A ball of mass 0.444 kg moving east (+c direction) with a speed of 3.90 m/s collides head- on with a 0.222 kg ball at rest. Assume that the collision is perfectly elastic. |v1=1 V | Units | atue Submit My Answers Give Up Part B What is be the direction of the velocity of the 0.444-kg ball after the collision? to the east to the west Submit My Answers Give Up Part C What is the speed of the 0.222-kg ball after the collision? Express your answer to three significant figures and include the appropriate units 1v2 =| V | Units | atue Submit My Answers Give Up

Explanation / Answer

Given

masses of balls  

m1 = 0.444 kg moving +x direction (east), moving with velocity u1 = 3.90 m/s

m2 = 0.222 kg at rest ==> u2 = 0 m/s

the collision is head-on and perfectly elastic so the conservation of both momentum and kinetic energy takes place and after the collision both ball are confined to the X axis only (head-on).

From the formula of final velocities of ball after the collision

Part A

v1 = ((m1-m2)/(m1+m2))u1

v1 = ((0.444-0.222)/(0.444+0.222))(3.90) m/s

v1 = 1.3 m/s

Part B

the final velocity of m1 is +ve so the direction is along the +x direction that is to the EAST

Part C

final speedof m2 after the collision is  

v2 = (2m1*u1)/(m1+M2) m/s

V2 = (2*0.444*3.9)/(0.444+0.222) m/s

v2 = 5.2 m/s

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