Answer:
a) 1.75m/s
b) 14,062.5J
Step-by-step explanation:
- a) We solve this problem using linear momentum
![p=mv](https://img.qammunity.org/2021/formulas/physics/high-school/6fwu51bwtbpnbnl3151q1laurhi2qezdog.png)
where
is the momentum,
is the mass, and
is the velocity.
the momentum of the first train before the collision:
![p_(1)=m_(1)v_(1)](https://img.qammunity.org/2021/formulas/physics/high-school/ns4fobpfbf0rm0zfjc84ppviepqj19iih7.png)
where
and
![v_(1)=2.5m/s](https://img.qammunity.org/2021/formulas/physics/high-school/3fkuwfuxkh4t51xzl6zq5zj0mbhn2aypvy.png)
![p_(1)=(25,000kg)(2.5m/s)](https://img.qammunity.org/2021/formulas/physics/high-school/v9wuyckto4v1v5ybqja0s51j64s18p3560.png)
![p_(1)=62,500kgm/s](https://img.qammunity.org/2021/formulas/physics/high-school/yatt9fg8xkv3lntqpqjfduzze1w9ihzka3.png)
the momentum of the second train before the collision:
![p_(2)=m_(2)v_(2)](https://img.qammunity.org/2021/formulas/physics/high-school/x6memfjy49cib2xg3xiyde63sehteh0jvn.png)
where
and
![v_(1)=1m/s](https://img.qammunity.org/2021/formulas/physics/high-school/vwly18k1g3w3ypi68vnzyci6ignxvx5hjd.png)
![p_(2)=(25,000kg)(1m/s)](https://img.qammunity.org/2021/formulas/physics/high-school/voy0zuvhr8ajtfhriymqd4tcxnh71rjegf.png)
![p_(2)=25,000kgm/s](https://img.qammunity.org/2021/formulas/physics/high-school/qy9oadn5boxyxxm08zh1wgfjzgo79xz5h7.png)
the total momentum before the collision is:
![p_(1)+p_(2)](https://img.qammunity.org/2021/formulas/physics/high-school/g9em2wpwh826u7dmc1fal3dxxdq3w3sn7k.png)
and due to the conservation of linear momentum: the amount of linear momentum before the collision must be the same after the collision.
so due to conservation:
![p_(1)+p_(2)=p_(f)](https://img.qammunity.org/2021/formulas/physics/high-school/8trs9b2h5bmnb41jcy1mm8kr3406yyby78.png)
the linear momentum of the two train system after the collision
![p_(f)=m_(f)v_(f)](https://img.qammunity.org/2021/formulas/physics/high-school/qxog6uaj71toy9waipx9ekzcwssne1xm7k.png)
where
is the final linear momentum,
is the final mass of the system, if the two cars end up connected the mass is:
(the sum of the mass of the two cars)
![m_(f)=50,000kg](https://img.qammunity.org/2021/formulas/physics/high-school/s5fw01ubpggd37tgrhnmb590ojv9c0iawr.png)
and
is the final speed: the speed of the connected cars.
so, going back to the conservation of momentum
![p_(1)+p_(2)=m_(f)v_(f)](https://img.qammunity.org/2021/formulas/physics/high-school/pa8tdoqptglcvl5nepfotxuudrx3f5p0es.png)
replacing all known values:
![87,500kgm/s=(50,000kg)v_(f)](https://img.qammunity.org/2021/formulas/physics/high-school/ulb45c0isxw8f64bs8119xij0subg8bh8d.png)
clearing for the final speed:
![v_(f)=(87,500kgm/s)/(50,000kg) =1.75m/s](https://img.qammunity.org/2021/formulas/physics/high-school/58bmazuwy2t7jndoasl4wa9gh3w92gjqjj.png)
- b) the initial kinetic energy:
![k1+k2](https://img.qammunity.org/2021/formulas/physics/high-school/4391zx65ca8nrrnmhtns8jlxljj3evbpgj.png)
which is:
![(1)/(2) m_(1)v_(1)^2+(1)/(2) m_(2)v_(2)](https://img.qammunity.org/2021/formulas/physics/high-school/ddarb1mv3jyoaunwpgho4wn0awoljdab1p.png)
replacing all known values:
![(1)/(2)(25,000kg)(2.5m/s)^2+(1)/(2)(25,000kg)(1m/s)^2\\=(1)/(2)(25,000kg)(6.25m^2/s^2)+(1)/(2)(25,000kg)(1m^2/s^2)\\=78,125J+12,500J\\=90,625J](https://img.qammunity.org/2021/formulas/physics/high-school/oezrt9ktaclvtk4k4lmlud6tmeh5o3bdwv.png)
anf the final kinetic energy is:
![k_(f)=(1)/(2) m_(f)v_(f)^2\\k_(f)=(1)/(2) (50,000kg)(1.75m/s)^2\\k_(f)=(1)/(2) (50,000kg)(3.0625m^2/s^2)\\k_(f)=76,562.5J](https://img.qammunity.org/2021/formulas/physics/high-school/eyqmk8dc73vfq4y8k3gtizqdknsueiidio.png)
the difference is:
![90,625J-76,562.5J=14,062.5J](https://img.qammunity.org/2021/formulas/physics/high-school/o3dr6tuuqihuuixyeb7rxbhls4f1xbyjlm.png)
the kinetic energy of the system decreased 14,062.5J