Determine the velocity of the block after the collision.

Question

A spring with a spring constant of 150N/m is compressed by a distance of 5 cm. A ball of mass 0.50 kg is placed at the end of the spring, and the spring is released and collides with the ball. The ball travels horizontally along a frictionless surface and ultimately collides with a block that has a mass of 0.75 kg in a perfectly elastic collision. Determine the velocity of the block after the collision.

Answer 1

So this problem has multiple physics concepts, Energy of a Spring, Conservation of Energy and finally conservation of momentum. First let's look at your knowns;

Part 1 Energy of a Spring

Given:

k = 150 N/m

Δx = 5 cm → 0.05 m ( need to convert to SI units)

Equations:

Potential Energy of a Spring = U_s = 1/2 k x²

Math:

U_s = 0.5 × 150 × 0.05² = 0.1875 J

We know that total mechanical energy is conserved. Therefore ME_{i =} ME_f Therefore the kinetic energy of the ball will have to equal the spring potential energy KE = Us

Given:

Us = 0.1875 J

m = 0.5 kg

Equation:

Us = KE = 1/2 m v²

Math

0.1875 = 1/2(.5) v²

v = √0.75

v = 0.866 m/s

We now have a conservation of momentum problem. The problem states the block is not moving and that they have a completely elastic collision.

Given

m_ball = 0.05 kg m_block = 0.75 kg

v_{ball i} = 0.866 m/s v_{block i} = 0 m/s

v_{ball f} = ? v_{block f} = ?

Find your initial momentum:

Equation

p = mv

p_i = p_ball + p_block = 0.05 × 0.866 + 0.75 × 0 = 0.433 kg m/s

pi = pf

0.433 = pf

The problem gives us the masses of the objects but not the velocities. We will have to set up a system of equations to solve. Because kinetic energy is conserved in a completely elastic collision we know that the KEi = KEf we know the mass stays the same and since the ratios are the same . We can reduce this relationship to

v_{ball i} + v_{block i} = v _{ball f} + v_{block f}

0.866 + 0 = -v _{ball f} + v_{block f} ( v _{ball f} is negative because it will bounce back in the opposite direction it was traveling)

Solving for Vball in terms of vblock give us

v _{ball f} = v_{block f} - 0.866 we will plug this into our conservation of momentum equation from above:

0.433 = .5 (v_{block f} - 0.866) +.75v_{block f}

0.866 = 1.25 v_{block f}

v_{block f} = 0.6928 m/s

wow this was a lot of physics!!!!

Answer 2

Hi Annet,

Dealing with springs is tricky (because the force exerted by the spring changes). Thus, you should always try to use conservation of energy. The energy of the compressed spring is transferred to the ball as kinetic energy. As for the last part, think about conservation of momentum. I see this question has already been answered but I wanted to provide some reasoning.

Answer 3

kx²/2 =- mv²/2; v = (kx²/m)^1/2 = (150N/m·0.0025m²/0.50kg)^1/2 = 0.866 m/s;

0.5·0.866 = 0.5·u + 0.75·w; 0.5·3/4/2 = 0.5u²/2 + 0.75w² and you have to find w