   Chapter 6, Problem 83AP

Chapter
Section
Textbook Problem

A 0.30-kg puck, initially at rest on a frictionless horizontal surface, is struck by a 0.20-kg puck that is initially moving along the x-axis with a velocity of 2.0 m/s. Alter the collision, the 0.20-kg puck has a speed of 1.0 m/s at an angle of θ = 53° to the positive x-axis. (a) Determine the velocity of the 0.30-kg puck after the collision, (b) Find the fraction of kinetic energy lost, in the collision.

(a)

To determine
The velocity of the 0.30kg puck after the collision.

Explanation

Given Info:

The initial velocity of the 0.20kg puck is 2.0m/s .

The 0.30kg puck is in rest initially

The final velocity of the 0.20kg puck is 1.0m/s and having an angle of 53.0° with respect to the x-axis.

Explanation:

The puck before and after the collision,

Consider 0.20kg puck as m1 and 0.30kg puck as m2 .

Consider the conservation of momentum in both the x direction,

(px)f=(px)i (I)

• (px)f is the final momentum in x direction
• (px)i is the initial momentum in x direction

Re-write the equation (I),

m1ufcos53°+m2vfcosθ=m1ui+0

• ui is the initial velocity of m1
• uf is the final velocity of m1
• vf is the final velocity of m2
• θ is the angle of m2 with x-axis

Re-write the above expression,

m2vfcosθ=m1uim1ufcos53.0° (II)

Consider the conservation of momentum in both the y direction,

(py)f=(py)i (III)

• (py)f is the final momentum in y direction
• (py)i is the initial momentum in y direction

Re-write the equation (III),

m1ufsin53°m2vfsinθ=0

• ui is the initial velocity of m1
• uf is the final velocity of m1
• vf is the final velocity of m2
• θ is the angle of m2 with x-axis

Re-write the above expression,

m2vfsinθ=m1ufsin53

(b)

To determine
The fraction of kinetic energy lost in the collision.

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