A 2.6 kg (mA) frictionless cart is moving at a constant speed of vA = 2.9 m/s to

Question

A 2.6 kg (mA) frictionless cart is moving at a constant speed of vA = 2.9 m/s to the right on a horizontal surface, as shown above, when it collides with a second cart of undetermined mass m that is initially at rest. The force F of the collision as a function of time t is shown in the graph below, where t = 0 is the instant of initial contact. As a result of the collision, the second cart acquires a speed of 1.5 m/s to the right. Assume that friction is negligible before, during, and after the collision. (a) Calculate the magnitude and direction of the velocity of the 2.6 kg cart after the collision. (b) Calculate the mass m of the second cart. **Look at pic for complete question** Show your work please

My Notes Ask Your Teacher -18 points mi A 2.6 kg (mA) frictionless cart is moving at a constant speed of VA 2.9 m/s to the right on a horizontal surface, as shown above, when it collides with a second cart of undetermined mass m that is initially at rest. The force F of the collision as a function of time t is shown in the graph below, where t 0 is the instant of initial contact. As a result of the collision, the second cart acquires a speed of 1.5 m/s to the right. Assume that friction is negligible before, during, and after the collision. Force versus Time Graph Force (kN) 4 0.2 0.4 0.6 0.8 Time (ms) This is a force vs. time graph. The function is in the shape of a symmetric triangle. The force starts at zero at time zero, and the peak on the graph is at 8kN at 0.5ms (a) Calculate the magnitude and direction of the velocity of the 2.6 kg cart after the collision. m/s (b) Calculate the mass m of the second cart. kg

Explanation / Answer

Before collision:

For mass A:

m = 2.6 kg

v = 2.9 m/s

for mass B

m = ?

v = 0

After collision:

The Impulse of the force = area under F-t curve = 0.5*1*8 = 4 Ns

Velocity of mass B, v = 1.5 m/s

hence change in mometum of block B = 1.5 m

Impulse = change iin momentum

1.5m = 4

m = 8/3 = 2.67 kg

for block A,

impulse = 4 Ns

change in momentum = 2.6*(v-2.9)

impulse = change in momentum,

2.6*(v-2.9) = 4

v = 4.44 m/s

a) magnitude = 4.44 m/s direction is to the right

b) mass of second cart m = 2.67 kg.

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