A 6.50-kg clay ball is thrown directly against a perpendicular brick wall at a v

Question

A 6.50-kg clay ball is thrown directly against a perpendicular brick wall at a velocity of 25.0 m/s and shatters into three pieces, which all fly backward, as shown in the figure. The wall exerts a force on the ball of 1,840. N for 0.100 s. One piece of mass 2.00 kg travels backward at a velocity of 10.5 m/s and an angle of 32.7° above the horizontal. A second piece of mass 1.00 kg travels at a velocity of 7.75 m/s and an angle of 27.3° below the horizontal What is the velocity of the third piece? 2.00 kg 25.0 m/s 27.3 6.50 kg 1.00 kg

Explanation / Answer

use the law of conservation of linear momentum

Initial momentum must be equal to terminal momentum at both x and y directions.

Pxi = Pxt and Pyi = Pyt

Pxi = m*Vi = 6.5*25=(+)162.5 kg*m/s

The impact of the wall = F*dt=1840*0.1=(-184 kg*m/s

Pxt =-184 + 162.5 = -21.5 -x direction)

At this point,the ball shatters in three pieces

m1= 2 kg , m2=1 kg , m3= 6.5 -3 = 3.5 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 *10.5*cos32.7 = (-17.67 kg*m/s

Px2=m2*V2*cos28=1*7.5 *cos27.3 = (-6.66 ) kg*m/s

Px3=Pxt - (Px1 + Px2) =

Px3 = -21.5 - (-17.67 -6.67) = (-)2.84 kg*m/s

Py1=m1*V1*sin32

Py1 =2*10.5*sin32.7 =(+)11.31 kg*m/s

Py2=m2*V2*sin27.3

Py2 =1*7.5*sin27.3 =(-)3.43 kg*m/s

Py3=Pyt - (Py1 + Py2)=0

Py3 = -(11.31-3.42 ) = - 7.89 kg*m/s

P3=((Px3)^2 + (Py3)^2)^1/2

P3 = 2.84^2 + 7.89^2

Pex = 8.38 kg*m/s

a) P3=m3*V3 ====> V3=8.48 /2= 4.19m/s

b) Py3/Px3 = tanx

tan x =2.84/7.89

tan theta = - 0.356

theta = -19.59 deg

Note: check for calclational errors

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