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# Determine the Speed of the Block After Collision?

A pendulum bob with a mass of m1=(8.70x10^0) kg is connected to a pivot point by a massless rod of length r=(1.300x10^1) cm, as shown in the figure. At the top of the pendulum the bob has a speed of v1=(1.500x10^0) m/s and rotates colliding with a block of mass m2=(1.3200x10^1) kg at the bottom. After the collision the bob has (-4.332x10^1) % of its speed at the bottom of the arc just before impact. If this percentage is negative the bob bounce backward. If the percentage is positive the bob continues in its original direction. Determine the speed of the block after the collision.

### 1 Answer

- NCSLv 72 months agoFavorite Answer
Both kinematics and conservation of energy lead to an impact speed of

v = √(2gh) = √(2g*2r) = √(2*9.8m/s²*2*0.1300m) = 2.257 m/s

conserve momentum:

8.70kg * 2.257m/s = 8.70kg * -0.4332 * 2.257m/s + 13.200kg*V

where V is the speed of the block after the collision.

V = 0.8433 m/s

Hope this helps!