The same physics student jumps off the back of her Laser again, but this time the Laser is

a) The student's speed after jumping is 1.07 m/s

b) The final speed of the laser is 10.4 m/s

Explanation:

a)

This issue can be approached through the momentum conservation principle: In the absence of external forces, the combined momentum of the student and the laser must remain unchanged. Hence, we can express:

where:

The initial momentum is zero

m = 42 kg signifies the mass of the laser

v = 1.5 m/s is the laser's final velocity

M = 59 kg is the mass of the student

V denotes the student's final velocity

Solving this for V, we can determine the student's speed:

Thus, the student's final speed calculates to 1.07 m/s.

b)

Here, both the laser and the student have a combined speed of 3.1 m/s prior to the student's jump; thus, the initial momentum isn't zero.

where:

m = 42 kg denotes the mass of the laser

M = 59 kg is the student’s mass

u = 3.1 m/s is their starting velocity

V = -2.1 m/s indicates the student's speed post-jump (she jumps backward)

v signifies the laser's final speed

When we resolve for v, we have:

Learn more about momentum:

m = 42kg
M = 59 kg
u = 3.1 m/s
V = -2.1 m/s
V=(m+M)u-Mv/m
=(42+59)(3.1)-(59)(-2.1)/42

10.4m/s