Question

Sally, who has a mass of 51 kg, is sledding at the park. She places her 6-kg sled at the top of a hill and then runs with a velocity of 2 m/s and jumps on it. What is the combined velocity of Sally and the sled right after she jumps on, in m/s? (Round your answer to one decimal place if necessary.)

Answer 1

Answer:

Combined velocity = 1.8m/s

Step-by-step explanation:

Since this is a momentum question, we will need to employ the principle of conservation of momentum which says that:

For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

Let m1 be the mass of Sally = 51kg,

m2 = mass of sled = 6kg,

initial velocity of sally before collision = 2m/s,

let the combined velocity of Sally and the sled after collision = vm/s

Momentum before collision is given as

$m_{1} u_{1} + m_{2} u_{2} = 51(2) + 6(0)\\= 102kgm/s$

initial velocity u2 of sled is zero because it was just placed at the top of the hill and not moving.

Momentum after collision i.e. when Sally jumps on the sled is given as:

$(m_{1} + m_{2})v = (51 + 6)v = 57v$

Equating the two expressions, we have:

$102 = 57v$

Divide both sides by 57

$\frac{102}{57} =\frac{57v}{57}\\ \\v = 1.7895m/s$ ≅ 1.8m/s (rounding up to one decimal place)

Therefore, the combined velocity of Sally and the sled after she jumps on it is 1.8m/s.