Question

During a snowball fight, two snowballs travelling towards each other collide head-on. The first is moving east at a speed of 15.3 m/s and has a mass of 0.450 kg. The second is moving west at 13.5 m/s. When the snowballs collide, they stick together and travel west at 3.50 meters per second. What is the mass of the second snowball?

A parent pushes a baby stroller from home to daycare along a level road with a force of 42 N directed at an angle of 30

Answer 1

FIRST QUESTION:

(1) Look at your given and unknown information. It makes things much clearer.

East is the positive direction.

v1,i = +15.3 m/s

m1 = 0.450kg

v2,i = -13.5 m/s

vf = -3.50 m/s

m2 = ?

(2) Review related formulas and rules.

Conservation of momentum:

   \(\Delta p=0\)   

Momentum:

   \(p =mv\)    

(3) Expand the conservation of momentum and set the initial and final conditions equal to each other.

   \(p_{i}=p_{f}\)    

(4) Substitue the equation of momentum into the conservation.

   \(m_{1}v_{1,i}+m_{2}v_{2,i}=m_{1}v_{1,f}+m_{2}v_{2,f}\)    

(5) Since both snowballs travel in the same direction after the collision, we can combine masses and treat the final velocities as one velocity.

   \(m_{1}v_{1,i}+m_{2}v_{2,i}=(m_{1}+m_{2})v_{f}\)    

(6) Substitute your values.

   \((0.450kg)(15.3\frac{m}{s})+(m_{2})(-13.5\frac{m}{s})=(0.450kg+m_{2})(-3.50\frac{m}{s})\)    

(7) Solve for m2.

   \((0.450kg)(15.3\frac{m}{s})-(0.450kg)(-3.50\frac{m}{s})=(-3.50\frac{m}{s})(m_{2})-(-13.5\frac{m}{s})(m_{2})\)    

   \(\frac{(0.450kg)(15.3+3.50\frac{m}{s})}{-3.50+13.5\frac{m}{s}}=m_{2}\)    

m2 = 0.846 kg

SECOND QUESTION

(1) Look at the given and unknown information.

F = 42N

Angle = 30 deg

displacement = 0.83 km = 830m

(2) Review the related equations.

   \(W = F(cos(\theta))\Delta x\)    

(3) Substitute your values.

      

HOPE THIS HELPS! :]