1. Answer the following about Mass and Momentum. a) A particle of mass m moving

Question

1. Answer the following about Mass and Momentum.

a) A particle of mass m moving at 10 m/s in the positive x direction makes a glancing elastic collision with a particle of mass 2m that's at rest before the collision. After the collision, m moves off at an angle of 45 o above the x-axis and 2 m moves off at 60o below the x-axis. What's the speed of the particle of mass 2m after the collision?

b) A 45 kg girl stands on a 7 kg wagon holding two 21 kg weights. She throws the weights horizontally off the back of the wagon at a speed of 8 m/s relative to herself. Assuming that the wagon was at rest initially, what's the speed of the girl relative to the ground after she throws both weights at the same time? If the wagon was initially at rest, what's the speed relative to the ground with which the girl will move after she throws the weights one at a time with a speed of 8 m/s each relative to herself?

Explanation / Answer

part a:

let speed of mass m after collision is v1 m/s and speed of mass 2m after collision be v2 m/s

as collision is elastic, momentum and kinetic energy will conserved.

conserving momentum along y axis:

initial momentum along y axis=final momentum along y axis

==>0=m*v1*sin(45)-2*m*v2*sin(60)

==> v1=2*v2*sin(60)/sin(45)=2.4495*v2....(1)

conserving momentum along x axis:

m*10=m*v1*cos(45)+2*m*v2*cos(60)

==>10=2.4495*v2*cos(45)+2*v2*cos(60)

==>v2=10/(2.4495*cos(45)+2*cos(60))=3.66 m/s

then v1=2.4495*v2=8.9658 m/s

so speed of the particle of mass 2m after collision is 3.66 m/s

part b:

let speed of the girl after throwing the weight be v m/s.

then conserving momentum ,

0=(21+21)*8-(45+7)*v

==>v=42*8/52=6.4615 m/s

so speed of the girl relative to the ground is 6.4615 m/s

let after throing first weight, speed of the wagon becomes v m/s

conserving momentum,

0=21*8-(21+45+7)*v

==>v=21*8/73=2.3014 m/s

let speed after throwing second weight, speed of the wagon be v1.

then conserving momentum:

(45+21+7)*2.3014=(45+7)*v1-21*(8+2.3014)

==>v1=(73*2.3014+21*10.3014)/52=7.391 m/s

so final speed will be 7.391 m/s

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