Question

A 70 kg woman and her 35 kg son are standing at rest on an ice rink, as shown. They push against each other for a time of 0.60 s, causing them to glide apart. The speed of the woman immediately after they separate is 0.55m/s. Assume that during the push, friction is negligible compared with the forces the people exert on each other.

 

(a) Calculate the initial speed of the son after the push. 

(b) Calculate the magnitude of the average force exerted on the son by the mother during the push. 

(c) How do the magnitude and direction of the average force exerted on the mother by the son during the push compare with those of the average force exerted on the son by the mother? Justify your answer. 

(d) After the initial push, the friction that the ice exerts cannot be considered negligible, and the mother comes to rest after moving a distance of 7.0 m across the ice. If their coefficients of friction are the same, how far does the son move after the push?

Answer 1

(a) Apply momentum conservation to the explosion 

p_before = 0 = p_after 0 = m_av_af + m_bv_bf 

0 = (70)(– 0.55) + (35)(v_bf) 

v_bf = 1.1 m/s 

b) J_son = ∆p_son F_on-son t = m(v_f – v_i) F(0.6) = (35)(0 – 1.1) = F = – 64 N

(c) Based on newtons third law action/reaction, the force on the son must be the same but in the opposite direction as the force on the mother. 

(d) On the son W_fk = ∆K – f_kd = ½ mv_f² – ½mv_i² – μmg d = ½m (0–v_i²) d = v_i²/(2μ) 

This would be the same formula for the mother’s motion with a different initial velocity. Since the mass cancels out we see the distance traveled is proportional to the velocity squared. The boy moves at twice the speed of the mother, so based on this relationship should travel 4x the distance. The mother traveled 7 m so the son would have a sliding distance of 28m.

(Alternatively, you could plug in the numbers for the mother to solve for μ and then plug in again using the same value of μ and the sons velocity to find the distance. μ is the same for both people.)