Question

1. A 70-kg swimmer dives horizontally off a 500-kg raft. The diver's speed immediately after leaving the raft is 6.0 m/s. A micro-sensor system attached to the edge of the raft measures the time interval during which the diver applies an impulse to the raft just prior to leaving the raft surface. If the time interval is read as 0.25 s, what is the magnitude of the average horizontal force by diver on the raft?

Answer 1

To solve this problem it is necessary to apply the concepts related to momentum theorem.

The equation for impulse is given as

$I = Ft$

Where

I = Force

t = Time

At the same time we have the equation for momentum is given as

$p = mv$

The impulse momentum theorem states that the change in momentum of an object is equal to the impulse applied to it. Therefore

I = p

Ft = mv

Solving to find the force

$F = \frac{mv}{t}$

$F = \frac{(70)(6)}{0.25}$

$F = 1680N$

Therefore the magnitude of the average horizontal force by diver on the raft is 1680N

Answer 2

The magnitude of the average horizontal force by the diver on the raft is 1,680N

The impulse applied by the diver is equal to the momentum of the raft

The for calculating impulse is expressed as:

I = Ft

F is the applied force

t is the time taken

Similarly, the equation for calculating the momentum of the raft is expressed as

$\rho = mv$

Equating both formulas to get the required average horizontal force;

$Ft = mv\\F=\frac{mv}{t}$

Substitute the given parameters into the formula to have:

m = 70kg

v = 6.0m/s

t = 0.25s

$F=\frac{70 \times 6}{0.25}\\F=\frac{420}{0.25}\\F= 1680N$

Hence the magnitude of the average horizontal force by the diver on the raft is 1,680N

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