Awagon is rolling forward on level ground. friction is negligible. the person sitting in the wagon is holding a rock. the total mass of the wagon, rider, and rock is 93.5 kg. the mass of the rock is 0.310 kg. initially the wagon is rolling forward at a speed of 0.540 m/s. then the person throws the rock with a speed of 16.5 m/s. both speeds are relative to the ground. find the speed of the wagon after the rock is thrown directly forward.

Explanation:

Given Data

Total mass=93.5 kg

Rock mass=0.310 kg

Initially wagon speed=0.540 m/s

rock speed=16.5 m/s

To Find

The speed of the wagon

Solution

As the wagon rolls, momentum is given as

P=mv

where

m is mass

v is speed

put the values

P=93.5kg × 0.540 m/s

P =50.49 kg×m/s

Now we have to find the momentum of rock

momentum of rock = mv

momentum of rock = (0.310kg)×(16.5 m/s)

momentum of rock =5.115 kg×m/s  

From the conservation of momentum we can find the wagons momentum So

wagon momentum=50.49 -5.115 = 45.375 kg×m/s  

Speed of wagon = wagon momentum/(total mass-rock mass)

Speed of wagon=45.375/(93.5-0.310)

Speed of wagon= 0.487 m/s

Throwing rock backward,

momentum of wagon = 50.49+5.115 = 55.605  kg×m/s

Speed of wagon = wagon momentum/(total mass-rock mass)

speed of wagon = 55.605  kg×m/s/(93.5kg-0.310kg)

speed of wagon= 0.5967 m/s