Question

Conservation of linear momentum

Answer 1

• m1=1000kg
• m2=8000kg
• V1=20m/s
• V2=15m/s

Let the velocity of truck after collision be y

Apply conservation of linear momentum

`\\ \tt\longmapsto m1v1+m2v2=(m1+m2)(v3)`

`\\ \tt\longmapsto 1000(20)+8000(15)=(1000+8000)(12+y)`

`\\ \tt\longmapsto 20000+120000=9000(12+x)`

`\\ \tt\longmapsto 140000=108000+9000x`

`\\ \tt\longmapsto 9000y=32000`

`\\ \tt\longmapsto 9y=32`

`\\ \tt\longmapsto y=32/9`

`\\ \tt\longmapsto y=3.5m/s`

#2

K.E(initial)

For car

`\\ \tt\longmapsto (1)/(2)(1000)(20)^2`

`\\ \tt\longmapsto 500(400)=20000J`

For truck

`\\ \tt\longmapsto (1)/(2)(8000)(15)^2`

`\\ \tt\longmapsto 4000(225)`

`\\ \tt\longmapsto 900000J`

• Total=920000J

K.E(Final)

`\\ \tt\longmapsto (1)/(2)(9000)(15.5)^2`

`\\ \tt\longmapsto 4500(240.25)`

`\\ \tt\longmapsto 1081125J`

Change:-

`\\ \tt\longmapsto 1081125-920000=161125J`