2Qs - Average Frictional Force & Velocity

[glasgowm]

I'm doing revision work from the beginning of the year and I can't find the formulas for these two questions. -

1) Calculate the velocity of the pellet before it strikes the putty


2) Calculate the average frictional force it takes for the car to come to rest if its traveling at 15ms and comes to a rest after 20m.

[Hootenanny]

It would be useful if you could quote the complete question(s) along with your attempts. Often one can find a formula for a particular situation by applying previous knowledge.

[glasgowm]

2) Two cars, travelling in the same direction, skid on a patch of smooth, level ice Car A, of mass 1400kg skids straight into the back of Car B of mass 1000kg.
The two cars join.
immediatly before the impact car B was moving with a speed of 8ms
Immediatly after the impact, both cars are moving with a speed of 15ms

I got the speed of car A before the impact to be 20ms using m1u1 + m2u2 = (m1m2)V

after a collision the cars leave the patch of ice and continue skidding along the road, the come to a rest in a distance of 20m after leaving the ice
Calculate the average frictional force

[Hootenanny]

2) Two cars, travelling in the same direction, skid on a patch of smooth, level ice Car A, of mass 1400kg skids straight into the back of Car B of mass 1000kg.
The two cars join.
immediatly before the impact car B was moving with a speed of 8ms
Immediatly after the impact, both cars are moving with a speed of 15ms

I got the speed of car A before the impact to be 20ms using m1u1 + m2u2 = (m1m2)V

Looks good to me.

after a collision the cars leave the patch of ice and continue skidding along the road, the come to a rest in a distance of 20m after leaving the ice
Calculate the average frictional force
HINT:Think about conservation of energy.

[Office_Shredder]

F=ma.

Since you know the mass, you're really looking for the average acceleration (decelleration in this case).

So what you need is an equation that, given distance, initial and final velocity, gives you acceleration.

[glasgowm]

uhh

A = v-u / t is the only one I can think of but that doesnt involve distance?

Here's what I done.

---
distance / speed
20 / 15
1.3333s = t
a = -15ms
F = ma
= 2400 * 15
= 36000N

[Office_Shredder]

uhh

A = v-u / t is the only one I can think of but that doesnt involve distance?

Have you ever seen the formula
D = (V2 - V02) / 2a

[Hootenanny]

Alternatively, one could consider the kinetic energy of the cars after collision and equate this with the work done by the average frictional force. Both methods are valid.