Linear Momentum conservation

1. In a pairs figure-skating competition, a 66 kg man and his 51 kg female partner stand facing each other on skates on the ice. If they push apart and the woman has a velocity of 1.5 m/s eastward, what is the magnitude of the velocity of her partner? (Neglect friction.) Express answer using two significant figures. B)What is the direction of the velocity of her partner?

2. A cherry bomb explodes into three pieces of equal mass. One piece has an initial velocity of 13 m/s x .Another piece has an initial velocity of 7.0 m/s x – 5.0 m/s y. What is the velocity of the third piece? Enter the x and y components of the velocity separated by a comma. Express your answer using two significant figures.

3. For a movie scene, a 76 kg stuntman drops from a tree onto a 51 kg sled that is moving on a frozen lake with a velocity of 13 m/s toward the shore. A) What is the speed of the sled after the stuntman is on board? B) If the sled hits the bank and stops, but the stuntman keeps on going, with what speed does he leave the sled? (Neglect friction.) Express answers using two significant figures.

4. At a county fair, two children ram each other head-on while riding on the bumper cars. Jill and her car, traveling left at 3.20 m/s , have a total mass of 340 kg. Jack and his car, traveling to the right at 1.90 m/s, have a total mass of 275 kg. A) Assuming the collision to be elastic, determine the speed of Jill and her car after the collision? B) Determine the direction of the velocity of Jill and her car after the collision? C)Determine the speed of Jack and his car after the collision? C)Determine the direction of the velocity of Jack and his car after the collision?

5. Two skaters with masses of 66 kg and 51 kg, respectively, stand 7.5 m apart, each holding one end of a piece of rope. A) If they pull themselves along the rope until they meet, how far does each skater travel? (Neglect friction.) B. If only the 51 kg skater pulls along the rope until she meets her friend (who just holds onto the rope), how far does each skater travel? Express answers using two significant figures separated by a comma.

1. The total momentum is the same. Initial they stand, total momentum is null

$0 = m_1*v_1+m_2*v_2$

$v_2 =-m_1/m_2*v_1 = -51*1.5/66 =-1.159 m/s =-1.16 m/s$

direction is to the west.

2. On the x axis

$m(v_{x1}+v_{x2}+v_{x3}) = 0$

$13 +7+v_{x3} =0$

$v_{x3} =-20 m/s$

on the y axis

$m(v_{y1}+v_{y2}+v_{y3}) =0$

$0-5+v_{y3} =0$

$v_{y3} =5 m/s$

the velocity of the third piece is $(-20,5) m/s$.

3. A) the total momentum is the same

$m*v_i = (m+M)*v_f$ , where $v_i$ is the initial speed and $v_f$ the final speed

$v_f = m*v_i/(m+M) =51*13/(51+76) =5.22 m/s$

B) again the total momentum is the same

$(m+M)*v_i =M*v_f$

$V_f =(m+M)*v_i/M = (51+76)*5.22/76 =8.72 m/s$

4. In an elastic collision the total momentum and the total kinetic energy is the same before and after the collision.

Let $v_1$, $v_3$ be the speed of jill before and after the collision, $m_1$ the mass of jill and her car

Let $v_2$, $v_4$ be the speed of jack before and after the collision , $m_2$ the mass of jack and his car.

One can write

$m_1*v_1+m_2*v_2 =m_1*v_3+m_2*v_4$

$m_1*v_1^2/2 + m_2*v_2^2/2 = m_1*v_3^2/2 + m_2*v_4^2/2$

One knows $m_a$, $m_2$, $v_1$ and $v_2$.

Solving the system of eq. for $v_3$ and $v_4$ one obtains

$v_3 =3.44 m/s$ and $v_4 =1.60 m/s$

direction of $v_3$ (jill after collision) is to the right

direction of $v_4$ (jack after collision) is to the left.

5. There is no difference in the forces exerted by one to the other skaters if one just stand and another pull the rope of if both skaters pull the rope. As the third principle states the force of action is equal to the force of reaction. This means that in both cases they will travel the same way. The distance traveled is proportional to their acceleration which is proportional to the force and to their mass. let $d_1$ be the distance traveled by $m_1$ (first skater) and $d_2$ the distance traveled by $m_2$ (the second skater)

$d_1+d_2 = d$

$d_1/d_2 = m_1/m_2$

$d_1*m_2=d_2*m_1$

$d_1 = m_1/m_2*d_2$

$d_2(1+m_1/m_2) = d$

$d_2 =d/(1+m_1/m_2) =7.5/(1+66/51) =3.27 m$

$d_1 =7.5-3.27 =4.23 m$