Question

3 Roller skating Your'e on a roller-skating rink with a friend who weighs twice as much as you. The two of you stand motionless in the middle of the rink so that your combined momentum is zero. You then push one another and begin to roll apart. (a) (5 points) If your momentum is then 450 N-s to the left, what is your friend's mo- mentum (include magnitude and direction)? (b) (10 points) If the push lasted for 2s, what is the average force with which you pushed? Hint: use the same method as in 2(b) (c) (10 points) After the push, what is the ratio between your friend's speed and your speed? Hint: either use variables m and 2m for the unknown masses, or simply assume a specific mass for yourself, eg. m = 80kg. The mass will cancel out when you form the ratio of speeds.

Answer 1

Let my weight be m my friends weight = 2m combined momentum = 0 = m_1 times v_1 + m_2 v_2 implies m_1v-1 = -m_2 v_2 given m_1 v_1 = momentum = 450N towards left = -450 N thus, - 450 N = -m_2 v_2 m_2 v-2 = P-2 = 450 N thus, my friends momentum is 450N towards right F-avg = dp/dt = 450/2 = 225 N My first speed P_2 = m_2 v-2 = mv_2 = 450 v-2 = 450/m my speed b_1 = m_1 v_1 = 2 mv_1 v_1 = 450/2M ratio = 450 /m/450/ 2m = 2: 1