Question

a boy and a cart are moving in the same direction, with the boy going twice as fast as the cart. when he gets into the cart the speed of the car increases by 20% . find the ratio of mass of cart to mass of boy

Answer 1

Answer:

The ratio of the mass of cart and mass of boy will be 4:1

Step-by-step explanation:

Conservation of momentum will be applied here,

Let the mass of the boy = x

let the mass of the cart = y

Let the speed of the cart be 100 m/s before

=> speed of the boy = 200 m/s

Hence momentum of the system before the boy gets into cart;

= 200x + 100y..............eq1

After the boy gets into the cart, effective mass = x + y

speed of the cart = 100 + 20% = 120 m/s

Hence net momentum of the system

= 120(x + y)

= 120 x + 120 y.....................eq2

since total momentum of the system will be conserved, hence equating eq1 and eq2 we get

200x + 100y = 120x + 120y

=> 80x = 20y

=> y/x = 80/20 = 4

Hence the ratio of the mass of cart and mass of boy will be 4:1

Answer 2

Answer:

M : m = 4:1

Step-by-step explanation:

As per given information suppose that

Mass of boy = m

let the mass of the cart = M

Speed of cart alone = v

Speed of boy alone = 2v

Speed of cart after boy gets into it = 1.2v

Using Law of conversation of momentum we know that momentum of system remains constant.

So, momentum of cart and boy before and after is equal

Momentum before = Momentum after

m(2v) + Mv = m(0) + (M+m)(1.2v)

2mv + Mv = 1.2Mv + 1.2mv

2mv - 1.2mv = 1.2Mv - Mv

0.8mv = 0.2Mv

8m = 2M

$\frac{M}{m}=\frac{8}{2}$

$\frac{M}{m}=\frac{4}{1}$

So, our ratio of masses is

M : m = 4:1