Question

a car of mass 1000kg travelling at 32m/s dashes into the rear of a truck of mass 8000kgmoving in the same direction with a velocity of 4m/s. after the collision the car bounces back with a velocity of 8m/s. what is the velocity of truck after the impact

Answer 1

Answer:

7 m/s

Explanation:

FOR THIS WE USE THE LAW OF CONSERVATION OF LINEAR MOMENTUM

M 1=MASS OF CAR 

M2=MASS OF TRUCK

U1=INITIAL VELOCITY OF CAR

U2=INITIAL VELOCITY OF TRUCK

V1=FINAL VELOCITY OF CAR

V2 FINAL VELOCITY OF TRUCK

CONSERVATION OF LINEAR MOMENTUM = 

M1U1+M2U2+M1V1+M2V2

HERE,

M1=1000KG

U1=32M/S

V1=8M/S

M2=8000M/S

U2=4M/S

V2=XM/S

HERE WE HAVE TO FIND X

PUTTING THE VALUES IN THE FORMULA

M1U1+M2U2=M1V1+M2V2

1000*32+8000*4=1000*8+8000X

32000+32000=8000+8000X

8000X=64000-8000

8000X=56000

X=56000/8000

X=7

THEREFORE V2 = 7M/S

Answer 2

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