A 6 kg clay ball is directly against a perpendicular brick wall at a velocity of

Question

A 6 kg clay ball is directly against a perpendicular brick wall at a velocity of 22 m/s and shatters Into three pieces, which all fly backward, as shown In the figure wall exerts a force on live ball of 2400 N for 0.1 s. One piece of mass 2 kg travels backward at a velocity of 17 m/s and an angle of 32 degree above the horizontal. A second piece of mass 1 kg travels at a velocity of 5 m/s and an angle of 28 degree below the horizontal. What is the velocity of the third piece? (Let up be the +y positive direction and to the right be the +x direction.)

Explanation / Answer

Given
mass= 6 kg
velocity = 22m/s
F = 2490 N
t = 0.1 s
mass = 2kg
v = 17 m/s
angle = 32 degrees
mass= 1 kg
velocity = 5 m/s
angle = 28 degrees

use Conservation of Momentum principles.
Initial momentum must be equal to terminal momentum at both x and y directions.

Pxi = Pxt and Pyi = Pyt
Pxi = m*Vi = 6*22 =( +)132 kg*m/s

The impact of the wall = F*dt=2490*0.1=(-)249 kg*m/s

Pxt=-249+132=(-)117 (It means -x direction)

At this point,the ball shatters in three peces.

m1= 2 kg , m2=1 kg , m3=3 kg

Total Px must be equal to Pxt and total Py must be equal to zero(Because there was no component of momentum vector in the first place)

Px1=m1*V1*cos32=2*17*cos32=(-)28.83 kg*m/s
Px2=m2*V2*cos28=1*5*cos28=(-)4.414 kg*m/s
Px3=Pxt - (Px1 + Px2) =-117-(-28.83-4.414)=(-)83.756 kg*m/s

Py1=m1*V1*sin32=(+)18.01 kg*m/s
Py2=m2*V2*sin28=(-)2.347 kg*m/s
Py3=Pyt - (Py1 + Py2)=0-(18.01-2.347)=(-)15.663 kg*m/s
P3=((Px3)^2 + (Py3)^2)^1/2 = (83.756^2+15.663^2)^0.5 = 85.21 kg*m/s
a) P3=m3*V3
85.21= 3*V3
V3 = 8.40 m/s
b) Py3/Px3 = tanx
tanx = (-15.663)/(-83.756) = 0.187
x= (-)10.6 dergrees below from the horizontal

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