Question

(a) A 100 gram golf ball is moving with a velocity of 20 m/s. It makes a head on perfectly elastic collision with an 8 kilogram steel ball, initially moving with a velocity of 1.2 m/s. Compute velocities of the balls after collision.
(b) Assume that the target is stationary (at rest) in the above problem. Can you estimate the final velocities (v1 and v2) of both objects without explicitly solving? In the above problem, m2>>m1, means m2 is very much heavier than m1.

Answer 1

Answer:

a)Golf ball= -17.13

Steel ball= 1.66

b) Golf ball= -20

Steel Ball=0

Explanation:

Hello There

a)

By using this equation

$V1= \frac{U1(M1-M2)+2(M2U2)}{(M1+M2)}$

and

$V2=\frac{U2(M2-M1)+2(M1U1)}{(M1+M2)}$.............(2)

final velocities can be calculated

(This Equation can be derived bu using conservation of momentum and energy. Its simple)

(If the first velocity is calculated the 2nd velocity can be calculated by using coefficient of restitution. As in elastic collision e=1)

So,

$V1=\frac{20(100-8000)+2(8000*1.2)}{(8100)}$............(1)

V1=-17.13

And

$V2=\frac{1.2(8000-100)+2(100*20)}{(8100)}$

V2=1.66

b)

By common sense we know that steel is much heavier than golf ball.

So a light object hitting a heavy body do noting much on the heavier one.

Rather than making the lighter one to retract.

Or you may use the previous equation and you will get an answer close to the initial velocity of lighter ball but with negative sign, meaning it retraces the same path

Hope this would help you a lot.

Please mark it as the brainliest.....

Answer 2

Answer:

pohello

the answer is

a)Golf ball= -17.13

Steel ball= 1.66

b) Golf ball= -20

Steel Ball=0

Explanation:

Hello There

a)

By using this equation

V1= \frac{U1(M1-M2)+2(M2U2)}{(M1+M2)}V1=

(M1+M2)

U1(M1−M2)+2(M2U2)

and

V2=\frac{U2(M2-M1)+2(M1U1)}{(M1+M2)}V2=

(M1+M2)

U2(M2−M1)+2(M1U1)

.............(2)

final velocities can be calculated

(This Equation can be derived bu using conservation of momentum and energy. Its simple)

(If the first velocity is calculated the 2nd velocity can be calculated by using coefficient of restitution. As in elastic collision e=1)

So,

V1=\frac{20(100-8000)+2(8000*1.2)}{(8100)}V1=

(8100)

20(100−8000)+2(8000∗1.2)

............(1)

V1=-17.13

And

V2=\frac{1.2(8000-100)+2(100*20)}{(8100)}V2=

(8100)

1.2(8000−100)+2(100∗20)

V2=1.66

b)

By common sense we know that steel is much heavier than golf ball.

So a light object hitting a heavy body do noting much on the heavier one.

Rather than making the lighter one to retract.

Or you may use the previous equation and you will get an answer close to the initial velocity of lighter ball but with negative sign, meaning it retraces the same path

Hope this would help you a lot.

Please mark it as the brainliest.....