An m = 6.60-kg clay ball is thrown directly against a perpendicular brick wall a

Question

An m = 6.60-kg clay ball is thrown directly against a perpendicular brick wall at a velocity of 22.0 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 2640 N for 0.110 s. One piece of mass m1 = 2.60 kg travels backward at a velocity of 10.5 m/s and an angle of = 32.0? above the horizontal. A second piece of mass m2 = 1.60 kg travels at a velocity of 8.50 m/s and an angle of 28.0? below the horizontal.

a) What is the velocity of the third piece?

b) What is the direction of the third piece?

Explanation / Answer

the impulse created by the wall on to the ball I

I= 2640*0.11290.4 N.s

using conservation of linear momentum along X and Y directions

let v3 be the velocity of 3rd component, be the angle of 3rd component to the horizontal

a) along X direction

mv+(m1v1cos32+m2v2cos28+m3v3cos)= I

6.6*22+(2.6*10.5cos32+1.6*8.5cos28+2.4*v3cos)=290.4

23.15+12+2.4*v3cos= 290.4- 145.2

v3cos=110.05/2.4=45.854

b) along Y direction

m1v1sin32-m2v2sin28-m3v3sin=0

2.6*10.5sin32-1.6*8.5sin28=2.4*v3sin

14.467-6.3848=2.4*v3sin

v3sin=3.3713

v3=sqrt[(v3sin)^2+ (v3cos)^2]

v3=45.98 m/sec

tan= vsin/vcos=0.073522

=4.2 degrees

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