# The Clam and Its Forces

1. Sep 16, 2008

### Phoenixtears

SOLVED

1. The problem statement, all variables and given/known data
Sea clams move by pushing water in one direction so that the water pushes them in the opposite direction. In doing so a 0.5 kg clam can accelerate 0.6 kg of water from rest to a speed of 1.9 m/s in 0.43 seconds. Calculate the magnitude of the clam's acceleration

2. Relevant equations
F= ma
Vf^2= V0^2 + 2ax
Vf= V0 + at
x= V0*t + .5a(t^2)

3. The attempt at a solution

I'm not sure why my answer isn't checking out. The initial velocity is 0. The final velocity is 1.9. The time is .43. These all fit into the third equation: 1.9= 0 + .43a. a then equals 4.42. This, however, is not working as my answer. What am I missing?

~Phoenix

Last edited: Sep 17, 2008
2. Sep 16, 2008

### ritwik06

Apply conservation of momentum
0.6*1.9=0.5*v
v=2.28
$$a=\frac{\Delta v}{\Delta t}$$
a=2.28/0.43
a=5.3
Does this help u?

3. Sep 17, 2008

### Ygggdrasil

The clam accelerates the water from 0 to 1.9 m/s in 0.43s. You need to calculate the acceleration of the clam, not the water. (hint: remember Newton's third law)

4. Sep 17, 2008

### Phoenixtears

OH! Alrighty, gotcha. That makes sense to me. Thank you so much. I realize now that I was mixing variables that shouldn't be used to solve for this.

Thank you so very much!