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 cart increases by20% .Find the ratio of mass of cart to mass of boy
Answers
Here, we use the concept of conservation of linear momentum, i.e., the total linear momentum of system of bodies is constant before and after the collision.
Because the boy gets into the cart, we assume it as a case of inelastic collision.
P1 + P2= P3
where P1= momentum of the boy
P2= momentum of the cart
P3= momentum of both
Let m and M be the mass of the boy and the cart with velocity 2v and v respectively.
P1= m*2v
P2= M*v
When the boy gets into the cart, its speed increases by 20%. It means the speed of the cart becomes 120% of initial velocity i.e. 1.2v
So,
P3= (m+M)*1.2v
Now,
Using the case of inelastic collision,
P1 + P2= P3
m*2v + M*v= (m+M)*1.2v
Taking v common from the left side,
v(2m + M)= (m+M)*1.2v
2m + M= (m + M)*1.2
2m + M= 1.2m + 1.2M
2m-1.2m= 1.2M-M
0.8*m= 0.2*M
m/M= 0.2/0.8
m/M= 1/4 or
M:m= 4:1
Therefore, the ratio of mass of cart to mass of boy is 4:1.
Here, we use the concept of conservation of linear momentum, i.e., the total linear momentum of system of bodies is constant before and after the collision.
Because the boy gets into the cart, we assume it as a case of inelastic collision.
P1 + P2= P3
where P1= momentum of the boy
P2= momentum of the cart
P3= momentum of both
Let m and M be the mass of the boy and the cart with velocity 2v and v respectively.
P1= m*2v
P2= M*v
When the boy gets into the cart, its speed increases by 20%. It means the speed of the cart becomes 120% of initial velocity i.e. 1.2v
So,
P3= (m+M)*1.2v
Now,
Using the case of inelastic collision,
P1 + P2= P3
m*2v + M*v= (m+M)*1.2v
Taking v common from the left side,
v(2m + M)= (m+M)*1.2v
2m + M= (m + M)*1.2
2m + M= 1.2m + 1.2M
2m-1.2m= 1.2M-M
0.8*m= 0.2*M
m/M= 0.2/0.8
m/M= 1/4 or
M:m= 4:1
Therefore, the ratio of mass of cart to mass of boy is 4:1.