A ball (mass = m ) moving in the positive x -direction with speed v collides wit

Question

A ball (mass = m) moving in the positive x-direction with speed v collides with a block (mass = M) initially at rest. Consider this an elastic collision.  After the collision, the block has a velocity of 6.0 m/s in the positive x-direction.  If m=3.0-kg and v=8.0 m/s, what is the mass of the block?

A ball (mass = m) moving in the positive x-direction with speed v collides with a block (mass = M) initially at rest. Consider this an elastic collision. After the collision, the block has a velocity of 6.0 m/s in the positive x-direction. If m=3.0-kg and v=8.0 m/s, what is the mass of the block?

Explanation / Answer

In an elastic collision, both momentum and kinetic energy are conserved.

m1v1i +m2v2i = m1v1f +m2v2f

and

m1v1i^2/2 +m2v2i^2/2 =m1v1f^2/2 +m2v2f^2/2

subbing in the things we do know...

(3.0kg)(8.0m/s) + M(0) = (3.0kg)v1f+ M(6.0m/s)

(3.0kg)(8.0m/s)^2/2 +M(0)^2/2 = (3.0kg)v1f^2/2 +M(6.0m/s)^2/2

simplifying...

(24.0kg m/s) = (3.0kg)v1f +M(6.0m/s) ... v1f = 8.0m/s - (2.0m/s kg)M

(96.0J) =(1.5kg)v1f2 +M(18.0m2/s2)

(96.0J) = (1.5kg)(8.0m/s - (2.0m/skg)M)^2 + M(18.0m2/s2)

(96.0) = (1.5)(64.0 - 32.0M +4.0M^2) + M(18.0)

(96.0) = (96.0 - 30.0M +6.0M^2)

using the quadratic equation,

you can solve for M = 0kg or M =5kg

M=5kg..........is answer

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