Erica (37kg ) and Danny (49kg ) are bouncing on a trampoline. Just as Erica reaches the high point of her bounce, Danny is moving upward past her at 4.3m/s . At that instant he grabs hold of her.

What is their speed just after he grabs her?

2.45 m/s

Explanation:

From the law of conservation of momentum,

The total momentum before the grab = Total momentum after the grab.

mu+m'u' = V(m+m') Equation 1

Where m = mass of Erica, m' = mass of Danny, u = initial velocity of Erica, u' = initial velocity of Danny, V = common velocity after garb

Make V the subject of the equation

V = (mu+m'u')/(m+m') Equation 2

Given: m = 37 kg, m' = 49 kg, u = 0 m/s(at the maximum height), u' = 4.3 m/s

Substitute into equation 2

V = (37×0+49×4.3)/(37+49)

V = 210.7/86

V = 2.45 m/s

Hence their speed just after the grab = 2.45 m/s

2.45 m/s

Explanation:

To solve this, we use formula for law of conservation of momentum, in which, no momentum is lost in the system.

The initial momentum is equal to the final momentum

m(i)v(i) = m(f)v(f)

Erica's momentum is zero, while Danny's momentum is 49 kg * 4.3 m/s

Danny's momentum = 210.7 kgm/s

Now, using law of conservation of momentum, we have

210.7 kgm/s = (37 + 49) kg * v(f)

210.7 kgm/s = 86 kg * v(f)

v(f) = 210.7 kgm/s / 86 kg

v(f) = 2.45 m/s

Therefore, their speed just after he grabs her is 2.45 m/s

change tension

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

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