# Category Archive: brakes

Sep 11

## A simple numerical on momentum

Car mass is 1200 kg moving velocity is 108 km/hr. On applying brakes velocity reduced to 36 km/hr. Find change in momentum?
(posted by Aditi)
Answer: m=1200 kg
u=108 x 5/18=30m/s
v=36×5/18=10m/s
Change in momentum = m(v-u) = 1200 x (30-10)=1200 ...

**Permanent link to this article: **http://www.plustwophysics.com/a-simple-numerical-on-momentum/

Sep 11

## A simple numerical on momentum

Car mass is 1200 kg moving velocity is 108 km/hr. On applying brakes velocity reduced to 36 km/hr. Find change in momentum?
(posted by Aditi)
Answer: m=1200 kg
u=108 x 5/18=30m/s
v=36×5/18=10m/s
Change in momentum = m(v-u) = 1200 x (30-10)=1200 ...

**Permanent link to this article: **http://www.plustwophysics.com/a-simple-numerical-on-momentum-2/

Sep 10

## Numerical Problem from Friction

“The friction coefficient between a road and the tyre of a vehicle is 4/3.find the maximum incline the road may have so that once hard brakes are applied and the wheel starts skidding,the vehicle going down at a speed of 10m/s is stopped within 5m”...

**Permanent link to this article: **http://www.plustwophysics.com/numerical-problem-from-friction/