Q8: Prb 5 - This problem has two parts to this. I will type out the questions for each of the two parts along with additional info in "Description" below.

Question

During an ice skating performance, an initially motionless 75 kg clown throws a fake barbell away horizontally. The clown's ice skates allow her to recoil frictionlessly.

Part a) If the clown recoils with a speed of 0.49 m/s and the barbell is thrown with a speed of 8.5 m/s, what is the mass, in kilograms, of the barbell?

Part b) How much kinetic energy, in joules, is gained by this maneuver?

Answer 1

Benson,

Part a) is solved using conservation of total linear momentum P_tot=m_cv_c+m_bv_b [bold-faced here and below are vectors, when they computer regular regular, it means we're using size (absolute value) of those vectors]

Part b) is solved using the KE definition

KE=mv²/2 [or KE=p²/2m ]

a) At start v_b=v_c=0 , therefore P_tot=0 . Because of conservation of total linear momentum, it will remain zero after the clown throws the barbel. At that moment

P_tot=m_cv_c+m_bv_b=0. Both v_{c and} v_b are horizontal ; they must be opposite to each other for P_tot=0

Therefore m_cv_c=m_bv_{b. .} From here

m_cv_c/v_{b=75kg.* ((0.49m/s)/8.5m/s)=4.3kg}

b) after the clone throws the barbell the total momentum of the system cologne plus barbell remains zero. That means that moment of barbel and the clown must be equal to each other. We could find that momentum

p=m_cv_c=75kg*0.49m/s=37kgm/s

Or p=m_bv_b =4.3kg*8.5m/s=37kgm/s

(only two significant.figs retained)

Knowing that I'm in them with allowed us to find the kinetic energy of the system after the cologne throws the barbell:

KE_totKE_c+KE_b=p²/2m_c+=p²/2m_b=p²*(1/2m_c+1/2m_b)=

(37kgm/s)²*[1/(2*75kg) + 1/(2*4.3kg)]=170J

Hope it is helpful

Dr.Ariel B.