A 4.0-kg object is moving with speed 2.0 m/s. A 1.0-kg object is moving with spe

Question

A 4.0-kg object is moving with speed 2.0 m/s. A 1.0-kg object is moving with speed 4.0 m/s. Both objects encounter the same constant braking force, and are brought to rest. Which object travels the greater distance before stopping? A 900-kg car travelling cast at 15.0 m/s suddenly collides with a 750-kg car traveling north at 20.0 m/s. The card stick together after the collision. What is the speed of the just after the collison? (Sketch the situation and Show all your work clearly for full credit) A 5.00-kg ball is hanging from a long but very light flexible wire when it is struck by a 1.50-kg stone traveling horizontally to the right at 1.20 m/s. The stone rebounds to the left with a speed of 8.50 m/s, and the ball swings to a maximum height h above its original level. The value of h is closest to

Explanation / Answer

12) constant braking force F is applied on both the masses

let a be the deceleration of the masses

a = F/m = F/4 for first object

a = F/m = F /1 for second object

v^2 - u^2 = 2aS

S = 2^2 / (2*F/4) = 8/F

S = 4^2 / (2*F) = 8/F

both will stop at the same distance

so option C) is the correct answer

13) let the velocity of the mass after collision be V m/s

conserving the momentum in x-direction we have

900*15 = (900+750)*v

Vx = 8.18 m/s

conserving the momentum in y - directionwe have

750*20 = (900+750)*v

Vy = 9.1 m/s

V = sqrt(Vx^2 + Vy^2)

V = sqrt(8.18^2 + 9.1^2) = 12.23 m/s

so option b is the correct answer

14) Ke energy given by the second ball = 0.5m(v1^2 - v2^2)

= 0.5*1.5*(12^2-8.5^2) = 53.8125 J

KE = PE of the hanging ball

53.8125 = mgh

h = 53.8125/5*9.8 = 1.0928 m

so option B) 1.1 m is the correct answer

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