Answer:
1.83m/s
Explanation:
According to the law of conservation of momentum which States that "the sum of momentum of bodies before collision is equal to the sum of momentum of the bodies after collision". The objects moves with a common velocity (v) after collision.
Since momentum = mass × velocity
Let m1 and m2 be the masses of the bag and the cart respectively and;
u1 and u2 be the velocities of the bag and the cart respectively.
v is the common velocity
momentum of the bag of rice before collision= m1u1
= 9×5.5
= 49.5kgm/s²
momentum of the cart before collision= m2u2
Since the grocery cart is stationary, its velocity will be 0m/s
Its momentum = 18(0) = 0kgm/s
Final momentum of the bodies after collision = (m1+m2)v
= (9+18)v
= 27v kgm/s
According to the law of conservation of momentum;
49.5+0 = 27v
v = 49.5/27
v = 1.83m/s
Therefore the final speed of the cart and bag is 1.83m/s
The speed with which the cart and bag move after the collision is equal to 1.83 m/s.
Given the following data:
To calculate the final speed with which the cart and bag move after the collision:
Applying the law of conservation of momentum, the collision between the cart and bag is given by the formula:
[tex]M_bV_b + M_cV_c = V_f(M_b + M_c)[/tex]
Where:
Substituting the given parameters into the formula, we have;
[tex]9.0 \times 5.5 + 18.0 \times 0 = V_f(9.0 + 18.0)\\\\49.5+0=V_f(27.0)\\\\49.5=27.0V_f\\\\V_f=\frac{49.5}{27.0} \\\\V_f=1.83 \;m/s[/tex]
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