A 6.65-kg bowling ball moving at 10.0 m/s collides with a 1.60-kg bowling pin, s

Question

A 6.65-kg bowling ball moving at 10.0 m/s collides with a 1.60-kg bowling pin, scattering it with a speed of 8.00 m/s and at an angle of 37.5° with respect to the initial direction of the bowling ball.

(a) Calculate the final velocity (magnitude and direction) of the bowling ball.

(b) Ignoring rotation, what was the original kinetic energy of the bowling ball before the collision?
J
(c) Ignoring rotation, what is the final kinetic energy of the system of the bowling ball and pin after the collision?
J

Explanation / Answer

let,

mass of the bowling ball, m1=6.65 kg

inital bowling ball velocity is u1=10 m/sec

final velocity is V=(v1X)i+(v1y)j

and

mass of the bowing pin m2=1.6 kg

final speed v2=8 m/sec at an angle of theta=37.5 degrrees,

then ,

v2x=v2*cos(theta)

v2y=v2*sin(theta)

now,

by using conservation of momemtum,

horizontal line

m1*u1+m2*v1=m2*v1x+m2*v2x

m1*u1+0=m2*v1x+m2*v2*cos(theta)

6.65*10+0=6.65*v1x+1.6*8*cos(37.5)

===> v1x=8.473 m/sec

and

verticle line

m1*u1+m2*v1=m2*v1y+m2*v2y

0+0=m2*v1y+m2*v2*sin(theta)

0=6.65*v1y+1.6*8*sin(37.5)

===> v1y=-1.172 m/sec

now,

final velocity of bowling ball is,

v1=sqrt(v1x^2+v1y^2)

v1=sqrt(8.473^2+(-1.172)^2)

v1=8.554 m/sec

and

tan(theta)=v1y/v1x

tan(theta)=-1.172/8.473

theta=-7.87 degrees

b)

kineteic energy is,

Ki=1/2*m1*u1^2

=1/2*6.65*10^2

=332.5 J

c)

Kf=1/2*m1*v1^2+1/2*m2*v2^2

=1/2*6.65*8.554^2+1/2*1.6*8^2

=194.49 J

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